{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261109-r05-binomial-theorem-0028",
  "pilot_number": "P250-0078",
  "number": 78,
  "family_id": "qf-mod40-binomial-expansion",
  "title": {
    "zh-Hant": "第 78 題：二項式定理",
    "en": "Question 78: The Finite Expansion Includes Every k from Zero through n"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261014-advance-20261109-r05-binomial-theorem-0028",
    "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-binomial-theorem"
    ],
    "title_zh": "二項式定理",
    "stem_zh": "commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={\"n\":2}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(2, [[0, 1, 2, 0], [1, 2, 1, 1], [2, 1, 0, 2]], complete_k_zero_through_n_coefficients)",
      "explanation_zh": "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。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "二項式定理"
      ]
    },
    "variant": {
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095028,
      "parameters": {
        "n": 2,
        "domain": "Exact_commutative_binomial_theorem",
        "operation": "binomial-expansion",
        "semantic_template": "qf-mod40-binomial-expansion/{\"n\":2}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須主端重算；不構成人工審定、single-element stabilizer、finite-prefix BW、single-cover compactness、zero-mass division、noncommutative binomial、incomplete Möbius factorization、bounded-search nonmembership、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-14",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261014-advance-20261109-r05-binomial-theorem-0028",
    "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "c00566b36326b791c4ab9313a94694fafa92b2e45d3b78f0935fd7f56275ac21",
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095028,
      "semantic_template": "qf-mod40-binomial-expansion/{\"n\":2}"
    },
    "fields": {
      "title": "The Finite Expansion Includes Every k from Zero through n",
      "stem": "Use natural exponent n=2, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary).",
      "answer": {
        "kind": "expression",
        "value": "(2, [[0, 1, 2, 0], [1, 2, 1, 1], [2, 1, 0, 2]], complete_k_zero_through_n_coefficients)",
        "explanation": "Every row is [k,C(n,k),x exponent n−k,y exponent k]. Retain all n+1 rows and exact integer coefficients. At n=0 the expansion is constant1, including the formal zero-exponent convention. Noncommuting variables require a different theorem. The exact result, retaining the requested order, is (2, [[0, 1, 2, 0], [1, 2, 1, 1], [2, 1, 0, 2]], complete_k_zero_through_n_coefficients)."
      },
      "skills": [
        "The Finite Expansion Includes Every k from Zero through n"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={\"n\":2}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use natural exponent n=2, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "n=2≥0; x,y commute",
      "after": "(x+y)^2 is a finite product of 2 factors",
      "relation": "implies",
      "rule_id": "finite_product",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個因子選 x 或 y，且乘法可交換，才能按 y 的選取次數合併同類項。",
        "en": "Choose x or y from each factor; commutativity permits grouping products by the number of y choices."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "此處不是非整數次方的無窮級數，也不套用到不可交換乘法。",
          "en": "This is not an infinite expansion for a noninteger power or a claim for noncommuting multiplication."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "choose exactly 0 of 2 factors to contribute y",
      "after": "C(2,0)=2/(1·2)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 2 個位置，再除掉 0 個 y 位置與 2 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 2 positions and divide by reorderings within the 0 y positions and 2 x positions, giving coefficient 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "2",
          "0"
        ],
        "output": "1",
        "label": "C(2,0)"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "C(2,0)=1; choose y 0 times",
      "after": "1·x^2·y^0",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 2-0=2；指數相加仍為 2。",
        "en": "The x exponent is 2-0=2; the two exponents still sum to 2."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "choose exactly 1 of 2 factors to contribute y",
      "after": "C(2,1)=2/(1·1)=2",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 2 個位置，再除掉 1 個 y 位置與 1 個 x 位置各自的重排，得到係數 2。",
        "en": "Count orderings of 2 positions and divide by reorderings within the 1 y positions and 1 x positions, giving coefficient 2."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "2",
          "1"
        ],
        "output": "2",
        "label": "C(2,1)"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "C(2,1)=2; choose y 1 times",
      "after": "2·x^1·y^1",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 2-1=1；指數相加仍為 2。",
        "en": "The x exponent is 2-1=1; the two exponents still sum to 2."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "choose exactly 2 of 2 factors to contribute y",
      "after": "C(2,2)=2/(2·1)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 2 個位置，再除掉 2 個 y 位置與 0 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 2 positions and divide by reorderings within the 2 y positions and 0 x positions, giving coefficient 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "2",
          "2"
        ],
        "output": "1",
        "label": "C(2,2)"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "C(2,2)=1; choose y 2 times",
      "after": "1·x^0·y^2",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 2-2=0；指數相加仍為 2。",
        "en": "The x exponent is 2-2=0; the two exponents still sum to 2."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "all k=0 through 2 have been accounted for",
      "after": "(x+y)^2=1·x^2·y^0 + 2·x^1·y^1 + 1·x^0·y^2",
      "relation": "implies",
      "rule_id": "collect_all_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一種 x/y 選取都落入唯一的 k 類，因此這是完整有限展開，不是只列頭尾項。",
        "en": "Every x/y choice belongs to exactly one k class, so this is the complete finite expansion, not just endpoint terms."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "共有 n+1 個係數位置，k 的全範圍逐一保留。",
          "en": "There are n+1 coefficient positions, with the complete range of k retained."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "n": 2,
    "terms": [
      [
        0,
        1,
        2,
        0
      ],
      [
        1,
        2,
        1,
        1
      ],
      [
        2,
        1,
        0,
        2
      ]
    ],
    "boundary": "complete_k_zero_through_n_coefficients"
  },
  "conclusion": {
    "canonical_value": "(2, [[0, 1, 2, 0], [1, 2, 1, 1], [2, 1, 0, 2]], complete_k_zero_through_n_coefficients)",
    "explanation": {
      "zh-Hant": "完整展開如上述所有單項式；欄序表逐項對應。",
      "en": "The full expansion consists of all displayed monomials, in the same order as the answer rows."
    }
  },
  "definition_note": {
    "zh-Hant": "n 是非負整數，x 與 y 可交換；n=0 使用多項式空乘積單位元。",
    "en": "n is a nonnegative integer and x,y commute. For n=0 use the polynomial empty-product identity."
  },
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      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
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    "scope_note": {
      "zh-Hant": "本稿是來源綁定的教學推導候選；精確重算不等於獨立數學審查或形式化證明。原題、答案、公式、AST 與歷史未被修改。",
      "en": "This is a source-bound pedagogical derivation candidate. Exact recomputation is not independent mathematical review or formal proof. Original questions, answers, formulas, ASTs and history are unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
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    "review_date": "2026-10-24",
    "owner_authorized_reader_publication": true,
    "scope": "Family prototypes and mathematical/bilingual reasoning variants reviewed in this existing context; parameter substitutions, all original/source bindings, full answer tuples and arithmetic ASTs independently reconstructed per question. Numeric binomial text slots are validated against independently regenerated n,k,C(n,k),n-k values. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
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    "formalization": "not_formalized",
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    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
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}