數列與級數/4 Sigma 的性質與求和公式-範例

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範例10

  • 求解：11^{3}+12^{3}+\cdots +20^{2}

答案

  • 41075

詳解

  • 11^{3}+12^{3}+\cdots +20^{3}=\sum\limits_{k=1}^{20}{k^{3}}-\sum\limits_{k=1}^{10}{k^{3}}
    =\left[ \frac{20\times (20+1)}{2} \right]^{2}-\left[ \frac{10\times (10+1)}{2} \right]^{2}=\left[ \frac{420}{2} \right]^{2}-\left[ \frac{110}{2} \right]^{2}=41075\#

範例10類題1

已知

求解

答案

詳解

範例10類題2

已知

求解

答案

詳解


範例11

  1. 求解：\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{n\times \left( n+1 \right)}
  2. 求解：\frac{1}{2^{2}-1}+\frac{1}{4^{2}-1}+\frac{1}{6^{2}-1}+\cdots +\frac{1}{100^{2}-1}


答案

  1. \frac{n}{n+1}
  2. \frac{50}{101}

詳解

1

  1. 求解：\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{n\times \left( n+1 \right)}
  2. \left\{ \begin{align} & \frac{1}{1\times 2}=\frac{1}{1}-\frac{1}{2} \\ & \frac{1}{2\times 3}=\frac{1}{2}-\frac{1}{3} \\ & \frac{1}{3\times 4}=\frac{1}{3}-\frac{1}{4} \\ & \cdots \\ & \frac{1}{\left( n+-11 \right)\times n}=\frac{1}{n-1}-\frac{1}{n} \\ & \frac{1}{n\times \left( n+1 \right)}=\frac{1}{n}-\frac{1}{n+1} \\ \end{align} \right.
  3. 各左式相加等於各右式相加，除第一式之正項及最後一式之負項外，右式相鄰兩式之正負項都可消去，故：
    \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{n\times \left( n+1 \right)}=\frac{1}{1}-\frac{1}{n+1}=\frac{n}{n+1}\#

2

  1. 求解：\frac{1}{2^{2}-1}+\frac{1}{4^{2}-1}+\frac{1}{6^{2}-1}+\cdots +\frac{1}{100^{2}-1}
  2. \left\{ \begin{align} & \frac{1}{2^{2}-1}=\frac{1}{\left( 2-1 \right)\times \left( 2+1 \right)}=\frac{1}{1\times 3}=\frac{1}{2}\left( \frac{1}{1}-\frac{1}{3} \right) \\ & \frac{1}{4^{2}-1}=\frac{1}{\left( 4-1 \right)\times \left( 4+1 \right)}=\frac{1}{3\times 5}=\frac{1}{2}\left( \frac{1}{3}-\frac{1}{5} \right) \\ & \frac{1}{6^{2}-1}=\frac{1}{\left( 6-1 \right)\times \left( 4+1 \right)}=\frac{1}{5\times 7}=\frac{1}{2}\left( \frac{1}{5}-\frac{1}{7} \right) \\ & \cdots \\ & \frac{1}{98^{2}-1}=\frac{1}{\left( 98-1 \right)\times \left( 98+1 \right)}=\frac{1}{97\times 99}=\frac{1}{2}\left( \frac{1}{97}-\frac{1}{99} \right) \\ & \frac{1}{100^{2}-1}=\frac{1}{\left( 100-1 \right)\times \left( 100+1 \right)}=\frac{1}{99\times 101}=\frac{1}{2}\left( \frac{1}{99}-\frac{1}{101} \right) \\ \end{align} \right.
  3. 各左式相加等於各右式相加，除第一式之正項及最後一式之負項外，右式相鄰兩式之正負項都可消去，故：
    \frac{1}{2^{2}-1}+\frac{1}{4^{2}-1}+\frac{1}{6^{2}-1}+\cdots +\frac{1}{100^{2}-1}=\frac{1}{2}\left( \frac{1}{1}-\frac{1}{101} \right)=\frac{1}{2}\times \frac{101-1}{101}=\frac{50}{101}\#

範例11類題

已知

求解

答案

詳解

取自"http://www.come2pass.com/wiki/index.php/%E6%95%B8%E5%88%97%E8%88%87%E7%B4%9A%E6%95%B8/4_Sigma_%E7%9A%84%E6%80%A7%E8%B3%AA%E8%88%87%E6%B1%82%E5%92%8C%E5%85%AC%E5%BC%8F-%E7%AF%84%E4%BE%8B"