次の3つの和を求める問題です。 (1) $\sum_{k=1}^{n} (5k+1)$ (2) $\sum_{k=1}^{n} (k+1)(k-2)$ (3) $\sum_{k=1}^{n} (k^3-k)$

1. 問題の内容

次の3つの和を求める問題です。

(1) $\sum_{k=1}^{n} (5k+1)$

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

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

2. 解き方の手順

(1)

$\sum_{k=1}^{n} (5k+1) = 5\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$

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

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

よって、

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

(2)

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

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

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

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

よって、

$\sum_{k=1}^{n} (k^2-k-2) = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} - 2n = \frac{2n^3+3n^2+n-3n^2-3n-12n}{6} = \frac{2n^3-14n}{6} = \frac{n^3-7n}{3}$

(3)

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

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

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

よって、

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

3. 最終的な答え

(1) $\frac{5n^2+7n}{2}$

(2) $\frac{n^3-7n}{3}$

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

または

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