以下の4つの和を求めます。 (1) $\sum_{k=1}^{n} (4k - 5)$ (2) $\sum_{k=1}^{n} (3k^2 - 7k + 4)$ (3) $\sum_{k=1}^{n} (k^3 + k)$ (4) $\sum_{k=1}^{n-1} 2k$

1. 問題の内容

以下の4つの和を求めます。

(1) $\sum_{k=1}^{n} (4k - 5)$

(2) $\sum_{k=1}^{n} (3k^2 - 7k + 4)$

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

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

2. 解き方の手順

(1)

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

(2)

$\sum_{k=1}^{n} (3k^2 - 7k + 4) = 3\sum_{k=1}^{n} k^2 - 7\sum_{k=1}^{n} k + \sum_{k=1}^{n} 4 = 3 \cdot \frac{n(n+1)(2n+1)}{6} - 7 \cdot \frac{n(n+1)}{2} + 4n = \frac{n(n+1)(2n+1)}{2} - \frac{7n(n+1)}{2} + 4n = \frac{n}{2}[(n+1)(2n+1) - 7(n+1) + 8] = \frac{n}{2}[2n^2 + 3n + 1 - 7n - 7 + 8] = \frac{n}{2}[2n^2 - 4n + 2] = n(n^2 - 2n + 1) = n(n-1)^2$

(3)

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

(4)

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

3. 最終的な答え

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

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

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

(4) $n^2 - n$