{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261103-r03-infinite-series-0068",
  "pilot_number": "R0034-17768",
  "number": 17768,
  "family_id": "qf-mod34-series-telescoping-sum",
  "title": {
    "zh-Hant": "第 17768 題：無窮級數",
    "en": "Question 17768: Retain Both the Telescoping Remainder and the Source Ratio Text"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261014-advance-20261103-r03-infinite-series-0068",
    "batch_id": "batch-20261014-advance-20261103-r03-infinite-series-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-infinite-series"
    ],
    "title_zh": "無窮級數",
    "stem_zh": "固定index convention的partial-sum sequence與exact tail。操作=series-telescoping-sum；資料={\"c\":18,\"shift\":2,\"n\":18}。請依 [c,shift,n,partial_sum,remainder,infinite_sum,certificate] 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(18, 2, 18, 81/10, 18/20, 18/2, exact_telescoping_remainder)",
      "explanation_zh": "Wave5第三層核對=(18, 2, 18, 81/10, 18/20, 18/2, exact_telescoping_remainder)；依完整明示資料重算；geometric要求絕對ratio小於1且telescoping保留remainder；項趨零必要但不充分；長有限prefix不證收斂。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 3,
      "skills": [
        "無窮級數"
      ]
    },
    "variant": {
      "family_id": "qf-mod34-series-telescoping-sum",
      "seed": 202611033068,
      "parameters": {
        "c": 18,
        "shift": 2,
        "n": 18,
        "domain": "Exact_symbolic_series_certificate",
        "operation": "series-telescoping-sum",
        "semantic_template": "qf-mod34-series-telescoping-sum/{\"c\":18,\"n\":18,\"shift\":2}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須由主端獨立重算；不構成人工審定、sampled rank upper bound、infinite-domain uniformity、finite-prefix series proof、posterior/identifiability、graph connectivity、nonexhaustive maximum、Lean或形式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-20261103-r03-infinite-series-0068",
    "batch_id": "batch-20261014-advance-20261103-r03-infinite-series-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "1dc091f10549be0a4472dda6ee1628713ee28a94e787d5b05c1e844a0cc56e82",
      "family_id": "qf-mod34-series-telescoping-sum",
      "seed": 202611033068,
      "semantic_template": "qf-mod34-series-telescoping-sum/{\"c\":18,\"n\":18,\"shift\":2}"
    },
    "fields": {
      "title": "Retain Both the Telescoping Remainder and the Source Ratio Text",
      "stem": "Use c=18, positive shift s=2, and n=18 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text.",
      "answer": {
        "kind": "expression",
        "value": "(18, 2, 18, 81/10, 18/20, 18/2, exact_telescoping_remainder)",
        "explanation": "Adjacent terms cancel, giving partial_sum=c/s−c/(s+n). The signed remainder tends to zero and the full sum is c/s. Keep the reduced computed partial sum and the original direct ratio representations as distinct output conventions; do not silently renormalize the answer bytes. The exact result, retaining the requested order, is (18, 2, 18, 81/10, 18/20, 18/2, exact_telescoping_remainder)."
      },
      "skills": [
        "Retain Both the Telescoping Remainder and the Source Ratio Text"
      ],
      "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": "固定index convention的partial-sum sequence與exact tail。操作=series-telescoping-sum；資料={\"c\":18,\"shift\":2,\"n\":18}。請依 [c,shift,n,partial_sum,remainder,infinite_sum,certificate] 欄序回答。",
        "en": "Use c=18, positive shift s=2, and n=18 terms c/(s+k)−c/(s+k+1) for k=0 through n−1. Give (c,shift,n,partial_sum,remainder,infinite_sum,exact_telescoping_remainder). The source directly formats remainder=c/(s+n) and infinite_sum=c/s and may leave those two ratios unreduced; retain that exact text."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "telescoping final remainder denominator := sum(2, 18)",
      "after": "telescoping final remainder denominator = 20",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留分母、正負號、索引與嚴格條件。",
        "en": "Compute exactly from original roles and established intermediates,retaining denominators,signs,indices and strict conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 telescoping final remainder denominator，不以小數近似替代。",
          "en": "Recompute telescoping final remainder denominator with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "2",
          "18"
        ],
        "output": "20",
        "label": "telescoping final remainder denominator"
      }
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "telescoping first endpoint numeric value := div(18, 2)",
      "after": "telescoping first endpoint numeric value = 9",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留分母、正負號、索引與嚴格條件。",
        "en": "Compute exactly from original roles and established intermediates,retaining denominators,signs,indices and strict conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 telescoping first endpoint numeric value，不以小數近似替代。",
          "en": "Recompute telescoping first endpoint numeric value with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "18",
          "2"
        ],
        "output": "9",
        "label": "telescoping first endpoint numeric value"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "telescoping signed remainder numeric value := div(18, 20)",
      "after": "telescoping signed remainder numeric value = 9/10",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留分母、正負號、索引與嚴格條件。",
        "en": "Compute exactly from original roles and established intermediates,retaining denominators,signs,indices and strict conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 telescoping signed remainder numeric value，不以小數近似替代。",
          "en": "Recompute telescoping signed remainder numeric value with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "18",
          "20"
        ],
        "output": "9/10",
        "label": "telescoping signed remainder numeric value"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "telescoping reduced partial sum := sub(9, 9/10)",
      "after": "telescoping reduced partial sum = 81/10",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留分母、正負號、索引與嚴格條件。",
        "en": "Compute exactly from original roles and established intermediates,retaining denominators,signs,indices and strict conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 telescoping reduced partial sum，不以小數近似替代。",
          "en": "Recompute telescoping reduced partial sum with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "9",
          "9/10"
        ],
        "output": "81/10",
        "label": "telescoping reduced partial sum"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "telescoping partial plus signed remainder := sum(81/10, 9/10)",
      "after": "telescoping partial plus signed remainder = 9",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "依原始角色及已核中間量精確計算，保留分母、正負號、索引與嚴格條件。",
        "en": "Compute exactly from original roles and established intermediates,retaining denominators,signs,indices and strict conditions."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 telescoping partial plus signed remainder，不以小數近似替代。",
          "en": "Recompute telescoping partial plus signed remainder with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "81/10",
          "9/10"
        ],
        "output": "9",
        "label": "telescoping partial plus signed remainder"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "original c=18;shift=2;n=18;terms c/(shift+k)-c/(shift+k+1),k0..n-1",
      "after": "ALL interior terms cancel;S_n=c/shift-c/(shift+n)=81/10;signed remainder=9/10;reverse sum=9",
      "relation": "implies",
      "rule_id": "series_telescoping_all_interior_cancellation_and_reverse_check",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個正的中間c/(s+k)與前一項的負值互消，留下首項與最後負項；加回帶符號餘項反核端點，不把負c的餘項改成絕對值。",
        "en": "Every interior positive endpoint cancels its preceding negative occurrence,leaving the first endpoint and final negative one.Add the signed remainder back to reverse-check;do not replace a negative-c remainder by its absolute value."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "same source formatting c=18;source remainder ratio text=18/20;source infinite-sum ratio text=18/2",
      "after": "computed partial sum REDUCED=81/10;remainder/infinite fields retain DIRECT UNREDUCED source text;numeric equivalents=[9/10, 9]",
      "relation": "implies",
      "rule_id": "series_telescoping_source_ratio_bytes_not_silent_reduction",
      "conditions": [],
      "explanation": {
        "zh-Hant": "來源只把計算出的partial sum約分，兩個直接ratio欄位可能仍為2/4或0/den；這是原輸出約定，保留字節並另核數值，不用數學等值當理由改原答案。",
        "en": "The source reduces the computed partial sum but directly formats the other two ratio fields,which can remain2/4 or0/den.Preserve those answer bytes and verify their numerical values separately;mathematical equivalence does not authorize rewriting the source output."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "DECLARED whole telescoping tail;remainder at general n is(18)/(n+2)",
      "after": "signed remainder tends to0;full partial-sum limit=c/shift under the given index convention",
      "relation": "implies",
      "rule_id": "series_telescoping_full_remainder_limit_not_one_n",
      "conditions": [],
      "explanation": {
        "zh-Hant": "固定c與正shift給完整餘項公式，n趨無限時才推出極限；不是只因當前n的端點核查成功。",
        "en": "The fixed coefficient and positive shift give an explicit remainder for every n.Its limit proves the full sum,not success of the single supplied endpoint check."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      18,
      2,
      18,
      "81/10",
      "18/20",
      "18/2",
      "exact_telescoping_remainder"
    ]
  },
  "conclusion": {
    "canonical_value": "(18, 2, 18, 81/10, 18/20, 18/2, exact_telescoping_remainder)",
    "explanation": {
      "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
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        "original_path": "public/data/mko/infinite-series.json",
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    "original_candidate_sha256": "ea6e0371f966deebeb0cf96f8ccdfa68bfa7597efa95482363a314b3253505af",
    "original_candidate_line_sha256": "ea6e0371f966deebeb0cf96f8ccdfa68bfa7597efa95482363a314b3253505af"
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      "formalization": "not_formalized",
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      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
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    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
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    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
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}