与えられた4つの数列の和を求める問題です。 (1) $\sum_{k=1}^{n} 6k^2$ (2) $\sum_{k=1}^{n} (2k^2+1)$ (3) $\sum_{k=1}^{n} k(3k-1)$ (4) $\sum_{k=1}^{n} (k-1)(k+2)$

1. 問題の内容

与えられた4つの数列の和を求める問題です。

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

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

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

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

2. 解き方の手順

数列の和の公式を利用して、それぞれの和を計算します。

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

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

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

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

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

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

(4) $\sum_{k=1}^{n} (k-1)(k+2) = \sum_{k=1}^{n} (k^2+k-2) = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k - \sum_{k=1}^{n} 2 = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} - 2n = \frac{n(n+1)(2n+1) + 3n(n+1) - 12n}{6} = \frac{n[(n+1)(2n+1) + 3(n+1) - 12]}{6} = \frac{n[2n^2+3n+1+3n+3-12]}{6} = \frac{n(2n^2+6n-8)}{6} = \frac{n(n^2+3n-4)}{3} = \frac{n(n+4)(n-1)}{3}$

3. 最終的な答え

(1) $n(n+1)(2n+1)$

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

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

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