次の数列の和 $S$ を求めます。 (1) $\frac{1}{1 \cdot 2 \cdot 3}, \frac{1}{2 \cdot 3 \cdot 4}, \frac{1}{3 \cdot 4 \cdot 5}, \dots, \frac{1}{n(n+1)(n+2)}$ (2) $\frac{1}{1 + \sqrt{3}}, \frac{1}{\sqrt{2} + \sqrt{4}}, \frac{1}{\sqrt{3} + \sqrt{5}}, \dots, \frac{1}{\sqrt{n} + \sqrt{n+2}}$

1. 問題の内容

次の数列の和 $S$ を求めます。

(1) $\frac{1}{1 \cdot 2 \cdot 3}, \frac{1}{2 \cdot 3 \cdot 4}, \frac{1}{3 \cdot 4 \cdot 5}, \dots, \frac{1}{n(n+1)(n+2)}$

(2) $\frac{1}{1 + \sqrt{3}}, \frac{1}{\sqrt{2} + \sqrt{4}}, \frac{1}{\sqrt{3} + \sqrt{5}}, \dots, \frac{1}{\sqrt{n} + \sqrt{n+2}}$

2. 解き方の手順

(1) 部分分数分解を利用します。$\frac{1}{n(n+1)(n+2)}$ を部分分数に分解すると、

$\frac{1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2}$

$1 = A(n+1)(n+2) + Bn(n+2) + Cn(n+1)$

$n=0$ のとき $1 = 2A$ より $A = \frac{1}{2}$

$n=-1$ のとき $1 = -B$ より $B = -1$

$n=-2$ のとき $1 = 2C$ より $C = \frac{1}{2}$

よって、

$\frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} \right) = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+1} - \frac{1}{n+1} + \frac{1}{n+2} \right) = \frac{1}{2} \left[ \left( \frac{1}{n} - \frac{1}{n+1} \right) - \left( \frac{1}{n+1} - \frac{1}{n+2} \right) \right]$

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

$S = \frac{1}{2} \left[ \left(1 - \frac{2}{2} + \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{2}{3} + \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{2}{4} + \frac{1}{5} \right) + \dots + \left( \frac{1}{n-2} - \frac{2}{n-1} + \frac{1}{n} \right) + \left( \frac{1}{n-1} - \frac{2}{n} + \frac{1}{n+1} \right) + \left( \frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} \right) \right]$

$S = \frac{1}{2} \left[ 1 - \frac{1}{2} - \frac{1}{n+1} + \frac{1}{n+2} \right] = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{n+1} + \frac{1}{n+2} \right] = \frac{1}{2} \left[ \frac{(n+1)(n+2) - 2(n+2) + 2(n+1)}{2(n+1)(n+2)} \right]$

$S = \frac{1}{2} \left[ \frac{n^2 + 3n + 2 - 2n - 4 + 2n + 2}{2(n+1)(n+2)} \right] = \frac{1}{2} \left[ \frac{n^2 + 3n}{2(n+1)(n+2)} \right] = \frac{n(n+3)}{4(n+1)(n+2)}$

(2) 分母の有理化を行います。

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

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

$S = \frac{1}{2} \left[ (\sqrt{3} - \sqrt{1}) + (\sqrt{4} - \sqrt{2}) + (\sqrt{5} - \sqrt{3}) + (\sqrt{6} - \sqrt{4}) + \dots + (\sqrt{n+1} - \sqrt{n-1}) + (\sqrt{n+2} - \sqrt{n}) \right]$

$S = \frac{1}{2} \left[ -\sqrt{1} - \sqrt{2} + \sqrt{n+1} + \sqrt{n+2} \right] = \frac{1}{2} \left[ \sqrt{n+1} + \sqrt{n+2} - 1 - \sqrt{2} \right]$

3. 最終的な答え

(1) $\frac{n(n+3)}{4(n+1)(n+2)}$

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