次の3つの和を求めます。 (1) $\sum_{k=1}^{n} k(k-1)$ (2) $1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots + n(2n+1)$ (3) $1^2 + 3^2 + 5^2 + \dots + (2n-1)^2$

1. 問題の内容

次の3つの和を求めます。

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

(2) $1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots + n(2n+1)$

(3) $1^2 + 3^2 + 5^2 + \dots + (2n-1)^2$

2. 解き方の手順

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

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

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

したがって、

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

(2) $1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots + n(2n+1) = \sum_{k=1}^{n} k(2k+1) = \sum_{k=1}^{n} (2k^2 + k) = 2\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$

$= 2\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} = \frac{2n(n+1)(2n+1) + 3n(n+1)}{6} = \frac{n(n+1)(2(2n+1)+3)}{6} = \frac{n(n+1)(4n+2+3)}{6} = \frac{n(n+1)(4n+5)}{6}$

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

$= 4\frac{n(n+1)(2n+1)}{6} - 4\frac{n(n+1)}{2} + n = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n = \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} = \frac{n(2(n+1)(2n+1) - 6(n+1) + 3)}{3} = \frac{n(2(2n^2+3n+1) - 6n - 6 + 3)}{3} = \frac{n(4n^2+6n+2-6n-3)}{3} = \frac{n(4n^2-1)}{3} = \frac{n(2n-1)(2n+1)}{3}$

3. 最終的な答え

(1) $\frac{n^3 - n}{3}$

(2) $\frac{n(n+1)(4n+5)}{6}$

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