等差級數n項和與2n項和及3n項和為等差數列之證明

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屬性

  • 資源類別：觀念
  • 科目：數學
  • 主題：數列與級數
  • 次主題：等差數列, 等差級數
  • 摘要：Sn，S2n-Sn，S3n-S2n，成等差數列
  • 適用年級：10-12
  • 日期：2008/10/13
  • 編輯者：User:HsiaoCH
  • 參考資料：
  • 相關技術：

  1. 已知：\left\{ \begin{align} & \left\{ a_{n} \right\}\to \text{an arithmetic sequence} \\ & S_{n}=\sum\limits_{k=1}^{n}{a_{k}},S_{2n}=\sum\limits_{k=1}^{2n}{a_{k}},S_{3n}=\sum\limits_{k=1}^{3n}{a_{k}} \\ \end{align} \right.
  2. 求證：S_{n},S_{2n}-S_{n},S_{3n}-S_{2n}\to \text{an arithmetic sequence}
  3. 依等差級數公式：\left\{ \begin{align} & S_{n}=\frac{n\times \left\{ \left[ a_{1} \right]+\left[ a_{1}+\left( n-1 \right)\times d \right] \right\}}{2}=\frac{n\times \left[ 2a_{1}+\left( n-1 \right)\times d \right]}{2} \\ & S_{2n}=\frac{2n\times \left\{ \left[ a_{1} \right]+\left[ a_{1}+\left( 2n-1 \right)\times d \right] \right\}}{2}=\frac{n\times \left[ 4a_{1}+2\left( 2n-1 \right)\times d \right]}{2} \\ & S_{3n}=\frac{3n\times \left\{ \left[ a_{1} \right]+\left[ a_{1}+\left( 3n-1 \right)\times d \right] \right\}}{2}=\frac{n\times \left[ 6a_{1}+3\left( 3n-1 \right)\times d \right]}{2} \\ \end{align} \right.
  4. S_{2n}-S_{n}=\frac{n\times \left[ 4a_{1}+2\left( 2n-1 \right)\times d \right]}{2}-\frac{n\times \left[ 2a_{1}+\left( n-1 \right)\times d \right]}{2}
    =\frac{n\times \left[ 4a_{1}+2\left( 2n-1 \right)\times d-2a_{1}-\left( n-1 \right)\times d \right]}{2}
    =\frac{n\times \left[ 2a_{1}+\left( 4n-2 \right)\times d-\left( n-1 \right)\times d \right]}{2}
    =\frac{n\times \left[ 2a_{1}+\left( 3n-1 \right)\times d \right]}{2}
  5. \left( S_{2n}-S_{n} \right)-S_{n}=\frac{n\times \left[ 2a_{1}+\left( 3n-1 \right)\times d \right]}{2}-\frac{n\times \left[ 2a_{1}+\left( n-1 \right)\times d \right]}{2}
    =\frac{n\times \left[ 2a_{1}+\left( 3n-1 \right)\times d-2a_{1}-\left( n-1 \right)\times d \right]}{2}
    =\frac{n\times \left( 2n\times d \right)}{2}={\color{Red}n^{2}d}
  6. S_{3n}-S_{2n}=\frac{n\times \left[ 6a_{1}+3\left( 3n-1 \right)\times d \right]}{2}-\frac{n\times \left[ 4a_{1}+2\left( 2n-1 \right)\times d \right]}{2}
    =\frac{n\times \left[ 6a_{1}+3\left( 3n-1 \right)\times d-4a_{1}-2\left( 2n-1 \right)\times d \right]}{2}
    =\frac{n\times \left[ 2a_{1}+\left( 9n-3 \right)\times d-\left( 4n-2 \right)\times d \right]}{2}
    =\frac{n\times \left[ 2a_{1}+\left( 5n-1 \right)\times d \right]}{2}
  7. \left( S_{3n}-S_{2n} \right)-\left( S_{2n}-S_{n} \right)=\frac{n\times \left[ 2a_{1}+\left( 5n-1 \right)\times d \right]}{2}-\frac{n\times \left[ 2a_{1}+\left( 3n-1 \right)\times d \right]}{2}
    =\frac{n\times \left[ 2a_{1}+\left( 5n-1 \right)\times d-2a_{1}-\left( 3n-1 \right)\times d \right]}{2}
    =\frac{n\times 2n\times d}{2}={\color{Red}n^{2}d}=\left( S_{2n}-S_{n} \right)-S_{n}
    \Rightarrow S_{n},S_{2n}-S_{n},S_{3n}-S_{2n}\to \text{an arithmetic sequence }\!\!\#\!\!\text{ }
Facts about 等差級數n項和與2n項和及3n項和為等差數列之證明RDF feed
主題 數列與級數  +
摘要 Sn，S2n-Sn，S3n-S2n，成等差數列  +
日期 2008年10月13日 (星期一)  +
次主題 等差數列  +, and 等差級數  +
科目 數學  +
編輯者 HsiaoCH  +
資源類別 觀念  +
適用年級 10-12  +
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