与えられた3つの和（Σ）の式を計算して、Σ記号を取り除いた形にします。 (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) $\sum_{k=1}^{n} \frac{3+2k}{k(k+1)(k+2)(k+3)}$

1. 問題の内容

与えられた3つの和（Σ）の式を計算して、Σ記号を取り除いた形にします。

(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。

$\frac{1}{k(k+1)(k+2)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2}$とおき、A, B, Cを求めます。

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

k=0のとき: $1 = 2A$ より $A = \frac{1}{2}$

k=-1のとき: $1 = -B$ より $B = -1$

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

よって、

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

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

$=\frac{1}{2} [(\frac{1}{1}-\frac{1}{2}) + (\frac{1}{2}-\frac{1}{3}) + ... + (\frac{1}{n} - \frac{1}{n+1}) - (\frac{1}{2}-\frac{1}{3}) - ... - (\frac{1}{n+1}-\frac{1}{n+2}) - (\frac{1}{1}-\frac{1}{2}) +(\frac{1}{n}-\frac{1}{n+1}) + (\frac{1}{n+1} - \frac{1}{n+2}) - (\frac{1}{2} - \frac{1}{3})- (\frac{1}{3} - \frac{1}{4})-... - (\frac{1}{n+1} - \frac{1}{n+2})]$

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

(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) 部分分数分解を行います。

$\frac{3+2k}{k(k+1)(k+2)(k+3)} = \frac{A}{k} + \frac{B}{k+1} + \frac{C}{k+2} + \frac{D}{k+3}$

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

k=0のとき: $3 = 6A$ より $A = \frac{1}{2}$

k=-1のとき: $1 = -2B$ より $B = -\frac{1}{2}$

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

k=-3のとき: $-3 = -6D$ より $D = \frac{1}{2}$

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

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

$\frac{1}{2} [(\frac{1}{1} - \frac{1}{2} - \frac{1}{3} + \frac{1}{4}) + (\frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5})+ ... + (\frac{1}{n} - \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3})] = \frac{1}{2} [(\frac{1}{1}-\frac{1}{2} - (\frac{1}{2}-\frac{1}{3}) - (\frac{1}{3}-\frac{1}{4})+...] = \frac{1}{2}[\frac{1}{1} - \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{4} +\frac{1}{5}-...\frac{1}{n}+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}]=\frac{1}{2} [\frac{1}{1} - (\frac{1}{n+1}+\frac{1}{n+2} - \frac{1}{n+2} - \frac{1}{n+1}) -\frac{1}{2}-\frac{1}{3} + \frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{2}-\frac{1}{3}]

$=\frac{1}{2} (\frac{1}{1} - \frac{1}{2} - \frac{1}{3} - \frac{1}{n+1} + \frac{1}{n+3})= \frac{1}{2} (\frac{1}{6} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+2} ) + \frac{1}{n+3}] = \frac{1}{2} (5/6 - \frac{1}{n+1} - \frac{1}{n+2}+\frac{1}{n+2}+\frac{1}{n+3} ]=\frac{1}{12} = \frac{5}{12} - (\frac{1}{n+1}+\frac{1}{n+1} ]- \frac{12} ( \frac{2}{6} - \frac{3n^2+5n]}{ = \frac{5}{6}- \frac{(n+2)(n+3)-(n+1)(n+3) + n^2+2 - n^3+2(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) $\sum_{k=1}^{n} \frac{3+2k}{k(k+1)(k+2)(k+3)} =\frac{5}{12} - \frac{1}{2(n+1)} - \frac{1}{2(n+2)} + \frac{1}{2(n+3)}$