以下の3つの和の式を計算します。 (1) $\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}$ (2) $\sum_{k=1}^{n} \frac{1}{\sqrt{k} + \sqrt{k+1}}$ (3) $\sum_{k=1}^{n} \frac{3+2k}{k(k+1)(k+2)(k+3)}$

はい、承知いたしました。画像の問題を解いていきます。

1. 問題の内容

以下の3つの和の式を計算します。

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

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

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

2. 解き方の手順

(1) 部分分数分解を利用します。

$\frac{1}{k(k+1)(k+2)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2}$ と置きます。

$1 = A(k+1)(k+2) + Bk(k+2) + Ck(k+1)$

$k=0$ のとき $1 = 2A$, よって $A = \frac{1}{2}$

$k=-1$ のとき $1 = -B$, よって $B = -1$

$k=-2$ のとき $1 = 2C$, よって $C = \frac{1}{2}$

よって、

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

したがって、

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

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

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

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

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

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

(2) 分母の有理化を行います。

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

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

$= \sqrt{n+1} - 1$

(3) 部分分数分解を利用します。

$\frac{3+2k}{k(k+1)(k+2)(k+3)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2} + \frac{D}{k+3}$ と置きます。

$3+2k = A(k+1)(k+2)(k+3) + Bk(k+2)(k+3) + Ck(k+1)(k+3) + Dk(k+1)(k+2)$

$k=0$ のとき $3 = 6A$, よって $A = \frac{1}{2}$

$k=-1$ のとき $1 = -2B$, よって $B = -\frac{1}{2}$

$k=-2$ のとき $-1 = 2C$, よって $C = -\frac{1}{2}$

$k=-3$ のとき $-3 = -6D$, よって $D = \frac{1}{2}$

よって、

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

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

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

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

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

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

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

3. 最終的な答え

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

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

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