$n$を2以上の整数とするとき、次の和$S_n$を求め、$n$の分数式で表す問題です。 $S_n = \sum_{k=2}^{n} \frac{k^2+1}{k^2-1}$

1. 問題の内容

$n$を2以上の整数とするとき、次の和$S_n$を求め、$n$の分数式で表す問題です。

$S_n = \sum_{k=2}^{n} \frac{k^2+1}{k^2-1}$

2. 解き方の手順

まず、$\frac{k^2+1}{k^2-1}$を部分分数分解します。

$\frac{k^2+1}{k^2-1} = \frac{k^2-1+2}{k^2-1} = 1 + \frac{2}{k^2-1} = 1 + \frac{2}{(k-1)(k+1)}$

$\frac{2}{(k-1)(k+1)}$を部分分数分解します。

$\frac{2}{(k-1)(k+1)} = \frac{A}{k-1} + \frac{B}{k+1}$とおくと、$2 = A(k+1) + B(k-1)$となります。

$k = 1$のとき、$2 = 2A$より、$A = 1$

$k = -1$のとき、$2 = -2B$より、$B = -1$

よって、$\frac{2}{(k-1)(k+1)} = \frac{1}{k-1} - \frac{1}{k+1}$

したがって、$\frac{k^2+1}{k^2-1} = 1 + \frac{1}{k-1} - \frac{1}{k+1}$

$S_n = \sum_{k=2}^{n} \frac{k^2+1}{k^2-1} = \sum_{k=2}^{n} (1 + \frac{1}{k-1} - \frac{1}{k+1}) = \sum_{k=2}^{n} 1 + \sum_{k=2}^{n} (\frac{1}{k-1} - \frac{1}{k+1})$

$\sum_{k=2}^{n} 1 = n-1$

$\sum_{k=2}^{n} (\frac{1}{k-1} - \frac{1}{k+1}) = (\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + \dots + (\frac{1}{n-3} - \frac{1}{n-1}) + (\frac{1}{n-2} - \frac{1}{n}) + (\frac{1}{n-1} - \frac{1}{n+1})$

$= 1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1} = \frac{3}{2} - \frac{n+1+n}{n(n+1)} = \frac{3}{2} - \frac{2n+1}{n(n+1)}$

$S_n = n-1 + \frac{3}{2} - \frac{2n+1}{n(n+1)} = n + \frac{1}{2} - \frac{2n+1}{n(n+1)} = \frac{2n^2(n+1) + n(n+1) - 2(2n+1)}{2n(n+1)} = \frac{2n^3+2n^2+n^2+n-4n-2}{2n(n+1)} = \frac{2n^3+3n^2-3n-2}{2n(n+1)}$

$2n^3+3n^2-3n-2 = (n-1)(2n^2+5n+2) = (n-1)(2n+1)(n+2)$より計算ミスがあります。

$\sum_{k=2}^n (\frac{1}{k-1} - \frac{1}{k+1}) = (\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + \cdots + (\frac{1}{n-1} - \frac{1}{n+1}) = 1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n+1} = \frac{3}{2} - \frac{2n+1}{n(n+1)} = \frac{3n^2+3n-4n-2}{2n(n+1)} = \frac{3n^2-n-2}{2n(n+1)} = \frac{(3n+2)(n-1)}{2n(n+1)}$

$S_n = n-1 + \frac{3n^2-n-2}{2n(n+1)} = \frac{2n^2(n+1) - 2n(n+1) + 3n^2-n-2}{2n(n+1)} = \frac{2n^3+2n^2-2n^2-2n+3n^2-n-2}{2n(n+1)} = \frac{2n^3+3n^2-3n-2}{2n(n+1)} = \frac{(n+2)(2n^2-n-1)}{2n(n+1)}$と書くこともできます。

$2n^3+3n^2-3n-2 = (2n+1)(n^2+n-2)=(2n+1)(n-1)(n+2)$.

したがって, $S_n = n-1+\frac{3}{2} - \frac{2n+1}{n(n+1)} = \frac{2n+1}{2}+\frac{n^2+n}{n(n+1)} = n+\frac{1}{2}-\frac{2n+1}{n(n+1)}$

$\sum_{k=2}^{n} \left(\frac{1}{k-1} - \frac{1}{k+1} \right) = \frac{3}{2}-\frac{1}{n} - \frac{1}{n+1} = \frac{3n^2+3n - (2n+2)}{2n(n+1)} = \frac{3n^2 + n -2 }{2n(n+1)} = \frac{(n-1)(3n+2)}{2n(n+1)}$

$S_n = (n-1)+\frac{(n-1)(3n+2)}{2n(n+1)}=\frac{2n(n+1)(n-1) + (n-1)(3n+2)}{2n(n+1)}=\frac{(n-1)[2n(n+1) + 3n+2]}{2n(n+1)}$

$=\frac{(n-1)(2n^2+5n+2)}{2n(n+1)}=\frac{(n-1)(n+2)(2n+1)}{2n(n+1)}$.

3. 最終的な答え

$S_n = \frac{(n-1)(n+2)(2n+1)}{2n(n+1)}$