与えられた3つの和の式について、それぞれの和を求める問題です。 (1) $\sum_{k=1}^{n} (2k - 2)$ (2) $\sum_{k=1}^{n} (4k^3 - 1)$ (3) $\sum_{k=1}^{n} (3k - 1)^2$

1. 問題の内容

与えられた3つの和の式について、それぞれの和を求める問題です。

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

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

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

2. 解き方の手順

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

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

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

$\sum_{k=1}^{n} 2 = 2n$

よって、

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

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

$\sum_{k=1}^{n} (4k^3 - 1) = 4 \sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} 1$

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

$\sum_{k=1}^{n} 1 = n$

よって、

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

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

$\sum_{k=1}^{n} (3k - 1)^2 = \sum_{k=1}^{n} (9k^2 - 6k + 1) = 9 \sum_{k=1}^{n} k^2 - 6 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$

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

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

$\sum_{k=1}^{n} 1 = n$

よって、

$9 \sum_{k=1}^{n} k^2 - 6 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 = 9 \cdot \frac{n(n+1)(2n+1)}{6} - 6 \cdot \frac{n(n+1)}{2} + n = \frac{3n(n+1)(2n+1)}{2} - 3n(n+1) + n = \frac{3n(n+1)(2n+1) - 6n(n+1) + 2n}{2} = \frac{n(3(n+1)(2n+1) - 6(n+1) + 2)}{2} = \frac{n(3(2n^2+3n+1) - 6n - 6 + 2)}{2} = \frac{n(6n^2 + 9n + 3 - 6n - 4)}{2} = \frac{n(6n^2 + 3n - 1)}{2}$

3. 最終的な答え

(1) $n(n-1)$

(2) $n(n^3 + 2n^2 + n - 1)$

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