次の数列の第n項までの和を求めます。 (1) $2^2, 4^2, 6^2, \dots$ (2) $1 \cdot 3, 2 \cdot 4, 3 \cdot 5, \dots$ (3) $1 \cdot 2 \cdot 3, 2 \cdot 3 \cdot 4, 3 \cdot 4 \cdot 5, 4 \cdot 5 \cdot 6, \dots$ (4) $\frac{1}{1 \cdot 3}, \frac{1}{2 \cdot 4}, \frac{1}{3 \cdot 5}, \frac{1}{4 \cdot 6}, \dots$

1. 問題の内容

次の数列の第n項までの和を求めます。

(1) $2^2, 4^2, 6^2, \dots$

(2) $1 \cdot 3, 2 \cdot 4, 3 \cdot 5, \dots$

(3) $1 \cdot 2 \cdot 3, 2 \cdot 3 \cdot 4, 3 \cdot 4 \cdot 5, 4 \cdot 5 \cdot 6, \dots$

(4) $\frac{1}{1 \cdot 3}, \frac{1}{2 \cdot 4}, \frac{1}{3 \cdot 5}, \frac{1}{4 \cdot 6}, \dots$

2. 解き方の手順

(1) 数列の一般項は $a_n = (2n)^2 = 4n^2$ です。

第n項までの和は

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} 4k^2 = 4 \sum_{k=1}^{n} k^2 = 4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$

(2) 数列の一般項は $a_n = n(n+2) = n^2 + 2n$ です。

第n項までの和は

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^2 + 2k) = \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{6} + n(n+1) = \frac{n(n+1)(2n+1) + 6n(n+1)}{6} = \frac{n(n+1)(2n+1+6)}{6} = \frac{n(n+1)(2n+7)}{6}$

(3) 数列の一般項は $a_n = n(n+1)(n+2) = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$ です。

第n項までの和は

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (k^3 + 3k^2 + 2k) = \sum_{k=1}^{n} k^3 + 3 \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k = \left(\frac{n(n+1)}{2}\right)^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1) = \frac{n^2(n+1)^2 + 2n(n+1)(2n+1) + 4n(n+1)}{4} = \frac{n(n+1)[n(n+1) + 2(2n+1) + 4]}{4} = \frac{n(n+1)[n^2+n + 4n + 2 + 4]}{4} = \frac{n(n+1)(n^2 + 5n + 6)}{4} = \frac{n(n+1)(n+2)(n+3)}{4}$

(4) 数列の一般項は $a_n = \frac{1}{n(n+2)} = \frac{1}{2} (\frac{1}{n} - \frac{1}{n+2})$ です。

第n項までの和は

$S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} \frac{1}{2} (\frac{1}{k} - \frac{1}{k+2}) = \frac{1}{2} \sum_{k=1}^{n} (\frac{1}{k} - \frac{1}{k+2}) = \frac{1}{2} [ (1 - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + \dots + (\frac{1}{n-1} - \frac{1}{n+1}) + (\frac{1}{n} - \frac{1}{n+2}) ] = \frac{1}{2} [ 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} ] = \frac{1}{2} [ \frac{3}{2} - \frac{2n+3}{(n+1)(n+2)} ] = \frac{1}{2} [ \frac{3(n+1)(n+2) - 2(2n+3)}{2(n+1)(n+2)} ] = \frac{3n^2 + 9n + 6 - 4n - 6}{4(n+1)(n+2)} = \frac{3n^2 + 5n}{4(n+1)(n+2)} = \frac{n(3n+5)}{4(n+1)(n+2)}$

3. 最終的な答え

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

(2) $\frac{n(n+1)(2n+7)}{6}$

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

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