# 二項式定理基礎 · 脈衝 R05

本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261014-advance-20261109-r05-binomial-theorem-foundations · 2026-10-14 · 100 題

欄序為來源綁定的呈現補充；原始 JSONL 與數學答案不變。

[欄序與來源綁定資料](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":0,"k":0}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(0, 0, 1, 1, 是)

Wave6第三層核對=(0, 0, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0001/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 2. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":1,"k":1}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(1, 1, 1, 1, 是)

Wave6第三層核對=(1, 1, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0002/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 3. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":2,"k":2}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(2, 2, 1, 1, 是)

Wave6第三層核對=(2, 2, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0003/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 4. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":3,"k":3}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(3, 3, 1, 1, 是)

Wave6第三層核對=(3, 3, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0004/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 5. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":4,"k":4}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(4, 4, 1, 1, 是)

Wave6第三層核對=(4, 4, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0005/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 6. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":5,"k":5}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(5, 5, 1, 1, 是)

Wave6第三層核對=(5, 5, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0006/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 7. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":6,"k":6}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(6, 6, 1, 1, 是)

Wave6第三層核對=(6, 6, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0007/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 8. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":7,"k":7}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(7, 7, 1, 1, 是)

Wave6第三層核對=(7, 7, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0008/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 9. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":8,"k":8}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(8, 8, 1, 1, 是)

Wave6第三層核對=(8, 8, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0009/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 10. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":9,"k":9}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(9, 9, 1, 1, 是)

Wave6第三層核對=(9, 9, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0010/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 11. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":10,"k":10}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(10, 10, 1, 1, 是)

Wave6第三層核對=(10, 10, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0011/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 12. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":11,"k":11}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(11, 11, 1, 1, 是)

Wave6第三層核對=(11, 11, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0012/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 13. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":12,"k":12}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(12, 12, 1, 1, 是)

Wave6第三層核對=(12, 12, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0013/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 14. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":13,"k":13}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(13, 13, 1, 1, 是)

Wave6第三層核對=(13, 13, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0014/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 15. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":14,"k":14}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(14, 14, 1, 1, 是)

Wave6第三層核對=(14, 14, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0015/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 16. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":15,"k":15}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(15, 15, 1, 1, 是)

Wave6第三層核對=(15, 15, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0016/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 17. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":16,"k":16}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(16, 16, 1, 1, 是)

Wave6第三層核對=(16, 16, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0017/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 18. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":17,"k":17}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(17, 17, 1, 1, 是)

Wave6第三層核對=(17, 17, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0018/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 19. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":18,"k":18}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(18, 18, 1, 1, 是)

Wave6第三層核對=(18, 18, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0019/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 20. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":19,"k":19}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(19, 19, 1, 1, 是)

Wave6第三層核對=(19, 19, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0020/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 21. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":20,"k":20}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(20, 20, 1, 1, 是)

Wave6第三層核對=(20, 20, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0021/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 22. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":21,"k":21}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(21, 21, 1, 1, 是)

Wave6第三層核對=(21, 21, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0022/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 23. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":22,"k":22}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(22, 22, 1, 1, 是)

Wave6第三層核對=(22, 22, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0023/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 24. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":23,"k":23}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(23, 23, 1, 1, 是)

Wave6第三層核對=(23, 23, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0024/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 25. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-coefficient；資料={"n":24,"k":24}。請依 [n, k, coefficient, symmetric_coefficient, equal] 欄序回答。

查看答案與解釋(24, 24, 1, 1, 是)

Wave6第三層核對=(24, 24, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0025/)

二項式定理 · qf-mod40-binomial-coefficient · candidate





## 26. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":0}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(0, [[0, 1, 0, 0]], constant_one_boundary)

Wave6第三層核對=(0, [[0, 1, 0, 0]], constant_one_boundary)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0026/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 27. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":1}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(1, [[0, 1, 1, 0], [1, 1, 0, 1]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(1, [[0, 1, 1, 0], [1, 1, 0, 1]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0027/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 28. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":2}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(2, [[0, 1, 2, 0], [1, 2, 1, 1], [2, 1, 0, 2]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(2, [[0, 1, 2, 0], [1, 2, 1, 1], [2, 1, 0, 2]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0028/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 29. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":3}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(3, [[0, 1, 3, 0], [1, 3, 2, 1], [2, 3, 1, 2], [3, 1, 0, 3]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(3, [[0, 1, 3, 0], [1, 3, 2, 1], [2, 3, 1, 2], [3, 1, 0, 3]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0029/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 30. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":4}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(4, [[0, 1, 4, 0], [1, 4, 3, 1], [2, 6, 2, 2], [3, 4, 1, 3], [4, 1, 0, 4]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(4, [[0, 1, 4, 0], [1, 4, 3, 1], [2, 6, 2, 2], [3, 4, 1, 3], [4, 1, 0, 4]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0030/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 31. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":5}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(5, [[0, 1, 5, 0], [1, 5, 4, 1], [2, 10, 3, 2], [3, 10, 2, 3], [4, 5, 1, 4], [5, 1, 0, 5]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(5, [[0, 1, 5, 0], [1, 5, 4, 1], [2, 10, 3, 2], [3, 10, 2, 3], [4, 5, 1, 4], [5, 1, 0, 5]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0031/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 32. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":6}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(6, [[0, 1, 6, 0], [1, 6, 5, 1], [2, 15, 4, 2], [3, 20, 3, 3], [4, 15, 2, 4], [5, 6, 1, 5], [6, 1, 0, 6]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(6, [[0, 1, 6, 0], [1, 6, 5, 1], [2, 15, 4, 2], [3, 20, 3, 3], [4, 15, 2, 4], [5, 6, 1, 5], [6, 1, 0, 6]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0032/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 33. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":7}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(7, [[0, 1, 7, 0], [1, 7, 6, 1], [2, 21, 5, 2], [3, 35, 4, 3], [4, 35, 3, 4], [5, 21, 2, 5], [6, 7, 1, 6], [7, 1, 0, 7]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(7, [[0, 1, 7, 0], [1, 7, 6, 1], [2, 21, 5, 2], [3, 35, 4, 3], [4, 35, 3, 4], [5, 21, 2, 5], [6, 7, 1, 6], [7, 1, 0, 7]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0033/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 34. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":8}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(8, [[0, 1, 8, 0], [1, 8, 7, 1], [2, 28, 6, 2], [3, 56, 5, 3], [4, 70, 4, 4], [5, 56, 3, 5], [6, 28, 2, 6], [7, 8, 1, 7], [8, 1, 0, 8]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(8, [[0, 1, 8, 0], [1, 8, 7, 1], [2, 28, 6, 2], [3, 56, 5, 3], [4, 70, 4, 4], [5, 56, 3, 5], [6, 28, 2, 6], [7, 8, 1, 7], [8, 1, 0, 8]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0034/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 35. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":9}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(9, [[0, 1, 9, 0], [1, 9, 8, 1], [2, 36, 7, 2], [3, 84, 6, 3], [4, 126, 5, 4], [5, 126, 4, 5], [6, 84, 3, 6], [7, 36, 2, 7], [8, 9, 1, 8], [9, 1, 0, 9]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(9, [[0, 1, 9, 0], [1, 9, 8, 1], [2, 36, 7, 2], [3, 84, 6, 3], [4, 126, 5, 4], [5, 126, 4, 5], [6, 84, 3, 6], [7, 36, 2, 7], [8, 9, 1, 8], [9, 1, 0, 9]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0035/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 36. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":10}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(10, [[0, 1, 10, 0], [1, 10, 9, 1], [2, 45, 8, 2], [3, 120, 7, 3], [4, 210, 6, 4], [5, 252, 5, 5], [6, 210, 4, 6], [7, 120, 3, 7], [8, 45, 2, 8], [9, 10, 1, 9], [10, 1, 0, 10]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(10, [[0, 1, 10, 0], [1, 10, 9, 1], [2, 45, 8, 2], [3, 120, 7, 3], [4, 210, 6, 4], [5, 252, 5, 5], [6, 210, 4, 6], [7, 120, 3, 7], [8, 45, 2, 8], [9, 10, 1, 9], [10, 1, 0, 10]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0036/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 37. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":11}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(11, [[0, 1, 11, 0], [1, 11, 10, 1], [2, 55, 9, 2], [3, 165, 8, 3], [4, 330, 7, 4], [5, 462, 6, 5], [6, 462, 5, 6], [7, 330, 4, 7], [8, 165, 3, 8], [9, 55, 2, 9], [10, 11, 1, 10], [11, 1, 0, 11]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(11, [[0, 1, 11, 0], [1, 11, 10, 1], [2, 55, 9, 2], [3, 165, 8, 3], [4, 330, 7, 4], [5, 462, 6, 5], [6, 462, 5, 6], [7, 330, 4, 7], [8, 165, 3, 8], [9, 55, 2, 9], [10, 11, 1, 10], [11, 1, 0, 11]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0037/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 38. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":12}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(12, [[0, 1, 12, 0], [1, 12, 11, 1], [2, 66, 10, 2], [3, 220, 9, 3], [4, 495, 8, 4], [5, 792, 7, 5], [6, 924, 6, 6], [7, 792, 5, 7], [8, 495, 4, 8], [9, 220, 3, 9], [10, 66, 2, 10], [11, 12, 1, 11], [12, 1, 0, 12]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(12, [[0, 1, 12, 0], [1, 12, 11, 1], [2, 66, 10, 2], [3, 220, 9, 3], [4, 495, 8, 4], [5, 792, 7, 5], [6, 924, 6, 6], [7, 792, 5, 7], [8, 495, 4, 8], [9, 220, 3, 9], [10, 66, 2, 10], [11, 12, 1, 11], [12, 1, 0, 12]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0038/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 39. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":13}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(13, [[0, 1, 13, 0], [1, 13, 12, 1], [2, 78, 11, 2], [3, 286, 10, 3], [4, 715, 9, 4], [5, 1287, 8, 5], [6, 1716, 7, 6], [7, 1716, 6, 7], [8, 1287, 5, 8], [9, 715, 4, 9], [10, 286, 3, 10], [11, 78, 2, 11], [12, 13, 1, 12], [13, 1, 0, 13]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(13, [[0, 1, 13, 0], [1, 13, 12, 1], [2, 78, 11, 2], [3, 286, 10, 3], [4, 715, 9, 4], [5, 1287, 8, 5], [6, 1716, 7, 6], [7, 1716, 6, 7], [8, 1287, 5, 8], [9, 715, 4, 9], [10, 286, 3, 10], [11, 78, 2, 11], [12, 13, 1, 12], [13, 1, 0, 13]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0039/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 40. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":14}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(14, [[0, 1, 14, 0], [1, 14, 13, 1], [2, 91, 12, 2], [3, 364, 11, 3], [4, 1001, 10, 4], [5, 2002, 9, 5], [6, 3003, 8, 6], [7, 3432, 7, 7], [8, 3003, 6, 8], [9, 2002, 5, 9], [10, 1001, 4, 10], [11, 364, 3, 11], [12, 91, 2, 12], [13, 14, 1, 13], [14, 1, 0, 14]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(14, [[0, 1, 14, 0], [1, 14, 13, 1], [2, 91, 12, 2], [3, 364, 11, 3], [4, 1001, 10, 4], [5, 2002, 9, 5], [6, 3003, 8, 6], [7, 3432, 7, 7], [8, 3003, 6, 8], [9, 2002, 5, 9], [10, 1001, 4, 10], [11, 364, 3, 11], [12, 91, 2, 12], [13, 14, 1, 13], [14, 1, 0, 14]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0040/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 41. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":15}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(15, [[0, 1, 15, 0], [1, 15, 14, 1], [2, 105, 13, 2], [3, 455, 12, 3], [4, 1365, 11, 4], [5, 3003, 10, 5], [6, 5005, 9, 6], [7, 6435, 8, 7], [8, 6435, 7, 8], [9, 5005, 6, 9], [10, 3003, 5, 10], [11, 1365, 4, 11], [12, 455, 3, 12], [13, 105, 2, 13], [14, 15, 1, 14], [15, 1, 0, 15]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(15, [[0, 1, 15, 0], [1, 15, 14, 1], [2, 105, 13, 2], [3, 455, 12, 3], [4, 1365, 11, 4], [5, 3003, 10, 5], [6, 5005, 9, 6], [7, 6435, 8, 7], [8, 6435, 7, 8], [9, 5005, 6, 9], [10, 3003, 5, 10], [11, 1365, 4, 11], [12, 455, 3, 12], [13, 105, 2, 13], [14, 15, 1, 14], [15, 1, 0, 15]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0041/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 42. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":16}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(16, [[0, 1, 16, 0], [1, 16, 15, 1], [2, 120, 14, 2], [3, 560, 13, 3], [4, 1820, 12, 4], [5, 4368, 11, 5], [6, 8008, 10, 6], [7, 11440, 9, 7], [8, 12870, 8, 8], [9, 11440, 7, 9], [10, 8008, 6, 10], [11, 4368, 5, 11], [12, 1820, 4, 12], [13, 560, 3, 13], [14, 120, 2, 14], [15, 16, 1, 15], [16, 1, 0, 16]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(16, [[0, 1, 16, 0], [1, 16, 15, 1], [2, 120, 14, 2], [3, 560, 13, 3], [4, 1820, 12, 4], [5, 4368, 11, 5], [6, 8008, 10, 6], [7, 11440, 9, 7], [8, 12870, 8, 8], [9, 11440, 7, 9], [10, 8008, 6, 10], [11, 4368, 5, 11], [12, 1820, 4, 12], [13, 560, 3, 13], [14, 120, 2, 14], [15, 16, 1, 15], [16, 1, 0, 16]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0042/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 43. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":17}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(17, [[0, 1, 17, 0], [1, 17, 16, 1], [2, 136, 15, 2], [3, 680, 14, 3], [4, 2380, 13, 4], [5, 6188, 12, 5], [6, 12376, 11, 6], [7, 19448, 10, 7], [8, 24310, 9, 8], [9, 24310, 8, 9], [10, 19448, 7, 10], [11, 12376, 6, 11], [12, 6188, 5, 12], [13, 2380, 4, 13], [14, 680, 3, 14], [15, 136, 2, 15], [16, 17, 1, 16], [17, 1, 0, 17]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(17, [[0, 1, 17, 0], [1, 17, 16, 1], [2, 136, 15, 2], [3, 680, 14, 3], [4, 2380, 13, 4], [5, 6188, 12, 5], [6, 12376, 11, 6], [7, 19448, 10, 7], [8, 24310, 9, 8], [9, 24310, 8, 9], [10, 19448, 7, 10], [11, 12376, 6, 11], [12, 6188, 5, 12], [13, 2380, 4, 13], [14, 680, 3, 14], [15, 136, 2, 15], [16, 17, 1, 16], [17, 1, 0, 17]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0043/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 44. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":18}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(18, [[0, 1, 18, 0], [1, 18, 17, 1], [2, 153, 16, 2], [3, 816, 15, 3], [4, 3060, 14, 4], [5, 8568, 13, 5], [6, 18564, 12, 6], [7, 31824, 11, 7], [8, 43758, 10, 8], [9, 48620, 9, 9], [10, 43758, 8, 10], [11, 31824, 7, 11], [12, 18564, 6, 12], [13, 8568, 5, 13], [14, 3060, 4, 14], [15, 816, 3, 15], [16, 153, 2, 16], [17, 18, 1, 17], [18, 1, 0, 18]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(18, [[0, 1, 18, 0], [1, 18, 17, 1], [2, 153, 16, 2], [3, 816, 15, 3], [4, 3060, 14, 4], [5, 8568, 13, 5], [6, 18564, 12, 6], [7, 31824, 11, 7], [8, 43758, 10, 8], [9, 48620, 9, 9], [10, 43758, 8, 10], [11, 31824, 7, 11], [12, 18564, 6, 12], [13, 8568, 5, 13], [14, 3060, 4, 14], [15, 816, 3, 15], [16, 153, 2, 16], [17, 18, 1, 17], [18, 1, 0, 18]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0044/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 45. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":19}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(19, [[0, 1, 19, 0], [1, 19, 18, 1], [2, 171, 17, 2], [3, 969, 16, 3], [4, 3876, 15, 4], [5, 11628, 14, 5], [6, 27132, 13, 6], [7, 50388, 12, 7], [8, 75582, 11, 8], [9, 92378, 10, 9], [10, 92378, 9, 10], [11, 75582, 8, 11], [12, 50388, 7, 12], [13, 27132, 6, 13], [14, 11628, 5, 14], [15, 3876, 4, 15], [16, 969, 3, 16], [17, 171, 2, 17], [18, 19, 1, 18], [19, 1, 0, 19]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(19, [[0, 1, 19, 0], [1, 19, 18, 1], [2, 171, 17, 2], [3, 969, 16, 3], [4, 3876, 15, 4], [5, 11628, 14, 5], [6, 27132, 13, 6], [7, 50388, 12, 7], [8, 75582, 11, 8], [9, 92378, 10, 9], [10, 92378, 9, 10], [11, 75582, 8, 11], [12, 50388, 7, 12], [13, 27132, 6, 13], [14, 11628, 5, 14], [15, 3876, 4, 15], [16, 969, 3, 16], [17, 171, 2, 17], [18, 19, 1, 18], [19, 1, 0, 19]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0045/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 46. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":20}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(20, [[0, 1, 20, 0], [1, 20, 19, 1], [2, 190, 18, 2], [3, 1140, 17, 3], [4, 4845, 16, 4], [5, 15504, 15, 5], [6, 38760, 14, 6], [7, 77520, 13, 7], [8, 125970, 12, 8], [9, 167960, 11, 9], [10, 184756, 10, 10], [11, 167960, 9, 11], [12, 125970, 8, 12], [13, 77520, 7, 13], [14, 38760, 6, 14], [15, 15504, 5, 15], [16, 4845, 4, 16], [17, 1140, 3, 17], [18, 190, 2, 18], [19, 20, 1, 19], [20, 1, 0, 20]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(20, [[0, 1, 20, 0], [1, 20, 19, 1], [2, 190, 18, 2], [3, 1140, 17, 3], [4, 4845, 16, 4], [5, 15504, 15, 5], [6, 38760, 14, 6], [7, 77520, 13, 7], [8, 125970, 12, 8], [9, 167960, 11, 9], [10, 184756, 10, 10], [11, 167960, 9, 11], [12, 125970, 8, 12], [13, 77520, 7, 13], [14, 38760, 6, 14], [15, 15504, 5, 15], [16, 4845, 4, 16], [17, 1140, 3, 17], [18, 190, 2, 18], [19, 20, 1, 19], [20, 1, 0, 20]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0046/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 47. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":21}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(21, [[0, 1, 21, 0], [1, 21, 20, 1], [2, 210, 19, 2], [3, 1330, 18, 3], [4, 5985, 17, 4], [5, 20349, 16, 5], [6, 54264, 15, 6], [7, 116280, 14, 7], [8, 203490, 13, 8], [9, 293930, 12, 9], [10, 352716, 11, 10], [11, 352716, 10, 11], [12, 293930, 9, 12], [13, 203490, 8, 13], [14, 116280, 7, 14], [15, 54264, 6, 15], [16, 20349, 5, 16], [17, 5985, 4, 17], [18, 1330, 3, 18], [19, 210, 2, 19], [20, 21, 1, 20], [21, 1, 0, 21]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(21, [[0, 1, 21, 0], [1, 21, 20, 1], [2, 210, 19, 2], [3, 1330, 18, 3], [4, 5985, 17, 4], [5, 20349, 16, 5], [6, 54264, 15, 6], [7, 116280, 14, 7], [8, 203490, 13, 8], [9, 293930, 12, 9], [10, 352716, 11, 10], [11, 352716, 10, 11], [12, 293930, 9, 12], [13, 203490, 8, 13], [14, 116280, 7, 14], [15, 54264, 6, 15], [16, 20349, 5, 16], [17, 5985, 4, 17], [18, 1330, 3, 18], [19, 210, 2, 19], [20, 21, 1, 20], [21, 1, 0, 21]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0047/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 48. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":22}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(22, [[0, 1, 22, 0], [1, 22, 21, 1], [2, 231, 20, 2], [3, 1540, 19, 3], [4, 7315, 18, 4], [5, 26334, 17, 5], [6, 74613, 16, 6], [7, 170544, 15, 7], [8, 319770, 14, 8], [9, 497420, 13, 9], [10, 646646, 12, 10], [11, 705432, 11, 11], [12, 646646, 10, 12], [13, 497420, 9, 13], [14, 319770, 8, 14], [15, 170544, 7, 15], [16, 74613, 6, 16], [17, 26334, 5, 17], [18, 7315, 4, 18], [19, 1540, 3, 19], [20, 231, 2, 20], [21, 22, 1, 21], [22, 1, 0, 22]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(22, [[0, 1, 22, 0], [1, 22, 21, 1], [2, 231, 20, 2], [3, 1540, 19, 3], [4, 7315, 18, 4], [5, 26334, 17, 5], [6, 74613, 16, 6], [7, 170544, 15, 7], [8, 319770, 14, 8], [9, 497420, 13, 9], [10, 646646, 12, 10], [11, 705432, 11, 11], [12, 646646, 10, 12], [13, 497420, 9, 13], [14, 319770, 8, 14], [15, 170544, 7, 15], [16, 74613, 6, 16], [17, 26334, 5, 17], [18, 7315, 4, 18], [19, 1540, 3, 19], [20, 231, 2, 20], [21, 22, 1, 21], [22, 1, 0, 22]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0048/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 49. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":23}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(23, [[0, 1, 23, 0], [1, 23, 22, 1], [2, 253, 21, 2], [3, 1771, 20, 3], [4, 8855, 19, 4], [5, 33649, 18, 5], [6, 100947, 17, 6], [7, 245157, 16, 7], [8, 490314, 15, 8], [9, 817190, 14, 9], [10, 1144066, 13, 10], [11, 1352078, 12, 11], [12, 1352078, 11, 12], [13, 1144066, 10, 13], [14, 817190, 9, 14], [15, 490314, 8, 15], [16, 245157, 7, 16], [17, 100947, 6, 17], [18, 33649, 5, 18], [19, 8855, 4, 19], [20, 1771, 3, 20], [21, 253, 2, 21], [22, 23, 1, 22], [23, 1, 0, 23]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(23, [[0, 1, 23, 0], [1, 23, 22, 1], [2, 253, 21, 2], [3, 1771, 20, 3], [4, 8855, 19, 4], [5, 33649, 18, 5], [6, 100947, 17, 6], [7, 245157, 16, 7], [8, 490314, 15, 8], [9, 817190, 14, 9], [10, 1144066, 13, 10], [11, 1352078, 12, 11], [12, 1352078, 11, 12], [13, 1144066, 10, 13], [14, 817190, 9, 14], [15, 490314, 8, 15], [16, 245157, 7, 16], [17, 100947, 6, 17], [18, 33649, 5, 18], [19, 8855, 4, 19], [20, 1771, 3, 20], [21, 253, 2, 21], [22, 23, 1, 22], [23, 1, 0, 23]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0049/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 50. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={"n":24}。請依 [n, coefficient_exponent_rows, boundary] 欄序回答。

查看答案與解釋(24, [[0, 1, 24, 0], [1, 24, 23, 1], [2, 276, 22, 2], [3, 2024, 21, 3], [4, 10626, 20, 4], [5, 42504, 19, 5], [6, 134596, 18, 6], [7, 346104, 17, 7], [8, 735471, 16, 8], [9, 1307504, 15, 9], [10, 1961256, 14, 10], [11, 2496144, 13, 11], [12, 2704156, 12, 12], [13, 2496144, 11, 13], [14, 1961256, 10, 14], [15, 1307504, 9, 15], [16, 735471, 8, 16], [17, 346104, 7, 17], [18, 134596, 6, 18], [19, 42504, 5, 19], [20, 10626, 4, 20], [21, 2024, 3, 21], [22, 276, 2, 22], [23, 24, 1, 23], [24, 1, 0, 24]], complete_k_zero_through_n_coefficients)

Wave6第三層核對=(24, [[0, 1, 24, 0], [1, 24, 23, 1], [2, 276, 22, 2], [3, 2024, 21, 3], [4, 10626, 20, 4], [5, 42504, 19, 5], [6, 134596, 18, 6], [7, 346104, 17, 7], [8, 735471, 16, 8], [9, 1307504, 15, 9], [10, 1961256, 14, 10], [11, 2496144, 13, 11], [12, 2704156, 12, 12], [13, 2496144, 11, 13], [14, 1961256, 10, 14], [15, 1307504, 9, 15], [16, 735471, 8, 16], [17, 346104, 7, 17], [18, 134596, 6, 18], [19, 42504, 5, 19], [20, 10626, 4, 20], [21, 2024, 3, 21], [22, 276, 2, 22], [23, 24, 1, 23], [24, 1, 0, 24]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0050/)

二項式定理 · qf-mod40-binomial-expansion · candidate





## 51. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":0,"x":-3,"y":2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(0, -3, 2, 1, 1, 是)

Wave6第三層核對=(0, -3, 2, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0051/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 52. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":1,"x":-2,"y":1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(1, -2, 1, -1, -1, 是)

Wave6第三層核對=(1, -2, 1, -1, -1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0052/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 53. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":2,"x":-1,"y":0}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(2, -1, 0, 1, 1, 是)

Wave6第三層核對=(2, -1, 0, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0053/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 54. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":3,"x":0,"y":-1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(3, 0, -1, -1, -1, 是)

Wave6第三層核對=(3, 0, -1, -1, -1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0054/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 55. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":4,"x":1,"y":-2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(4, 1, -2, 1, 1, 是)

Wave6第三層核對=(4, 1, -2, 1, 1, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0055/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 56. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":5,"x":2,"y":2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(5, 2, 2, 1024, 1024, 是)

Wave6第三層核對=(5, 2, 2, 1024, 1024, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0056/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 57. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":6,"x":3,"y":1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(6, 3, 1, 4096, 4096, 是)

Wave6第三層核對=(6, 3, 1, 4096, 4096, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0057/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 58. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":7,"x":-3,"y":0}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(7, -3, 0, -2187, -2187, 是)

Wave6第三層核對=(7, -3, 0, -2187, -2187, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0058/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 59. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":8,"x":-2,"y":-1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(8, -2, -1, 6561, 6561, 是)

Wave6第三層核對=(8, -2, -1, 6561, 6561, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0059/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 60. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":9,"x":-1,"y":-2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(9, -1, -2, -19683, -19683, 是)

Wave6第三層核對=(9, -1, -2, -19683, -19683, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0060/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 61. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":10,"x":0,"y":2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(10, 0, 2, 1024, 1024, 是)

Wave6第三層核對=(10, 0, 2, 1024, 1024, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0061/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 62. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":11,"x":1,"y":1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(11, 1, 1, 2048, 2048, 是)

Wave6第三層核對=(11, 1, 1, 2048, 2048, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0062/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 63. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":12,"x":2,"y":0}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(12, 2, 0, 4096, 4096, 是)

Wave6第三層核對=(12, 2, 0, 4096, 4096, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0063/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 64. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":13,"x":3,"y":-1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(13, 3, -1, 8192, 8192, 是)

Wave6第三層核對=(13, 3, -1, 8192, 8192, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0064/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 65. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":14,"x":-3,"y":-2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(14, -3, -2, 6103515625, 6103515625, 是)

Wave6第三層核對=(14, -3, -2, 6103515625, 6103515625, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0065/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 66. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":15,"x":-2,"y":2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(15, -2, 2, 0, 0, 是)

Wave6第三層核對=(15, -2, 2, 0, 0, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0066/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 67. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":16,"x":-1,"y":1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(16, -1, 1, 0, 0, 是)

Wave6第三層核對=(16, -1, 1, 0, 0, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0067/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 68. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":17,"x":0,"y":0}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(17, 0, 0, 0, 0, 是)

Wave6第三層核對=(17, 0, 0, 0, 0, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0068/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 69. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":18,"x":1,"y":-1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(18, 1, -1, 0, 0, 是)

Wave6第三層核對=(18, 1, -1, 0, 0, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0069/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 70. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":19,"x":2,"y":-2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(19, 2, -2, 0, 0, 是)

Wave6第三層核對=(19, 2, -2, 0, 0, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0070/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 71. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":20,"x":3,"y":2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(20, 3, 2, 95367431640625, 95367431640625, 是)

Wave6第三層核對=(20, 3, 2, 95367431640625, 95367431640625, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0071/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 72. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":21,"x":-3,"y":1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(21, -3, 1, -2097152, -2097152, 是)

Wave6第三層核對=(21, -3, 1, -2097152, -2097152, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0072/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 73. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":22,"x":-2,"y":0}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(22, -2, 0, 4194304, 4194304, 是)

Wave6第三層核對=(22, -2, 0, 4194304, 4194304, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0073/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 74. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":23,"x":-1,"y":-1}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(23, -1, -1, -8388608, -8388608, 是)

Wave6第三層核對=(23, -1, -1, -8388608, -8388608, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0074/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 75. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-evaluation-identity；資料={"n":24,"x":0,"y":-2}。請依 [n, x, y, power, expansion_sum, equal] 欄序回答。

查看答案與解釋(24, 0, -2, 16777216, 16777216, 是)

Wave6第三層核對=(24, 0, -2, 16777216, 16777216, 是)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0075/)

二項式定理 · qf-mod40-binomial-evaluation-identity · candidate





## 76. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-1}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-1, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-1, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0076/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 77. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-2}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-2, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-2, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0077/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 78. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-3}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-3, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-3, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0078/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 79. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-4}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-4, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-4, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0079/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 80. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-5}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-5, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-5, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0080/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 81. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-6}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-6, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-6, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0081/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 82. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-7}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-7, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-7, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0082/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 83. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-8}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-8, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-8, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0083/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 84. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-9}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-9, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-9, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0084/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 85. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-10}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-10, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-10, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0085/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 86. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-11}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-11, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-11, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0086/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 87. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-12}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-12, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-12, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0087/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 88. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-13}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-13, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-13, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0088/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 89. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-14}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-14, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-14, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0089/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 90. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-15}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-15, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-15, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0090/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 91. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-16}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-16, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-16, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0091/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 92. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-17}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-17, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-17, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0092/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 93. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-18}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-18, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-18, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0093/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 94. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-19}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-19, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-19, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0094/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 95. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-20}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-20, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-20, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0095/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 96. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-21}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-21, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-21, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0096/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 97. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-22}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-22, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-22, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0097/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 98. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-23}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-23, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-23, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0098/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 99. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-24}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-24, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-24, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0099/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate





## 100. 二項式定理

commutative coefficient setting與n in natural including zero。操作=binomial-exponent-boundary；資料={"exponent":-25}。請依 [exponent, finite_polynomial_claim, boundary] 欄序回答。

查看答案與解釋(-25, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)

Wave6第三層核對=(-25, 否, finite_binomial_polynomial_requires_natural_zero_exponent_and_commutative_multiplication)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261109-r05-binomial-theorem-0100/)

二項式定理 · qf-mod40-binomial-exponent-boundary · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261109-r05-binomial-theorem-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261109-r05-binomial-theorem-foundations/validation-report.json)