160112 初版 160112 更新
\(\displaystyle{S=\sum_{k=1}^n \dfrac{1}{(2k-1)(2k+1)}}\) とおく。
H君の方法
(一般論 分数の数列の和と不定方程式)
\(S=\left(1-\dfrac{2}{3}\right)\)
\(+\left(\dfrac{2}{3}-\dfrac{3}{5}\right)\)
\(+\left(\dfrac{3}{5}-\dfrac{4}{7}\right)\)
\(+\left(\dfrac{4}{7}-\dfrac{5}{9}\right)\)
\(+\cdots+\left(\dfrac{n}{2n-1}-\dfrac{n+1}{2n+1}\right)\)
\(=1-\dfrac{n+1}{2n+1}\)
ちょっと感動!
\((2n+1)n-(2n-1)(n+1)=1\)
| \(\dfrac{1}{1\cdot 3}\) |
= |
\(\dfrac{1}{1}\) |
- |
\(\dfrac{2}{3}\) |
| \(\dfrac{1}{3\cdot 5}\) |
= |
\(\dfrac{2}{3}\) |
- |
\(\dfrac{3}{5}\) |
| \(\dfrac{1}{5\cdot 7}\) |
= |
\(\dfrac{3}{5}\) |
- |
\(\dfrac{4}{7}\) |
| \(\dfrac{1}{7\cdot 9}\) |
= |
\(\dfrac{4}{7}\) |
- |
\(\dfrac{5}{9}\) |