次の和を求める問題です。 (1) $\sum_{k=1}^{n} (2k+1)$ (2) $\sum_{k=1}^{n} (k^2 + 5k)$ (3) $\sum_{k=1}^{n} (k^2 + 3k - 1)$ (4) $\sum_{k=1}^{n} (2k+1)^3$ (5) $\sum_{k=1}^{n} (k-1)(2k+3)$

1. 問題の内容

次の和を求める問題です。

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

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

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

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

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

2. 解き方の手順

(1)

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

(2)

$\sum_{k=1}^{n} (k^2 + 5k) = \sum_{k=1}^{n} k^2 + 5\sum_{k=1}^{n} k = \frac{n(n+1)(2n+1)}{6} + 5 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1) + 15n(n+1)}{6} = \frac{n(n+1)(2n+1+15)}{6} = \frac{n(n+1)(2n+16)}{6} = \frac{2n(n+1)(n+8)}{6} = \frac{n(n+1)(n+8)}{3}$

(3)

$\sum_{k=1}^{n} (k^2 + 3k - 1) = \sum_{k=1}^{n} k^2 + 3\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = \frac{n(n+1)(2n+1)}{6} + 3 \cdot \frac{n(n+1)}{2} - n = \frac{n(n+1)(2n+1) + 9n(n+1) - 6n}{6} = \frac{n(n+1)(2n+1+9) - 6n}{6} = \frac{n(n+1)(2n+10) - 6n}{6} = \frac{2n(n+1)(n+5) - 6n}{6} = \frac{n(2(n+1)(n+5) - 6)}{6} = \frac{n(2(n^2 + 6n + 5) - 6)}{6} = \frac{n(2n^2 + 12n + 10 - 6)}{6} = \frac{n(2n^2 + 12n + 4)}{6} = \frac{2n(n^2 + 6n + 2)}{6} = \frac{n(n^2 + 6n + 2)}{3}$

(4)

$\sum_{k=1}^{n} (2k+1)^3 = \sum_{k=1}^{n} (8k^3 + 12k^2 + 6k + 1) = 8\sum_{k=1}^{n} k^3 + 12\sum_{k=1}^{n} k^2 + 6\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 = 8(\frac{n(n+1)}{2})^2 + 12\frac{n(n+1)(2n+1)}{6} + 6\frac{n(n+1)}{2} + n = 8\frac{n^2(n+1)^2}{4} + 2n(n+1)(2n+1) + 3n(n+1) + n = 2n^2(n+1)^2 + 2n(n+1)(2n+1) + 3n(n+1) + n = 2n^2(n^2+2n+1) + 2n(2n^2+3n+1) + 3n^2 + 3n + n = 2n^4+4n^3+2n^2 + 4n^3 + 6n^2 + 2n + 3n^2 + 4n = 2n^4 + 8n^3 + 11n^2 + 6n = n(2n^3 + 8n^2 + 11n + 6)$

(5)

$\sum_{k=1}^{n} (k-1)(2k+3) = \sum_{k=1}^{n} (2k^2 + 3k - 2k - 3) = \sum_{k=1}^{n} (2k^2 + k - 3) = 2\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k - 3\sum_{k=1}^{n} 1 = 2\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} - 3n = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} - 3n = \frac{2n(n+1)(2n+1) + 3n(n+1) - 18n}{6} = \frac{n(2(n+1)(2n+1) + 3(n+1) - 18)}{6} = \frac{n(2(2n^2+3n+1) + 3n + 3 - 18)}{6} = \frac{n(4n^2+6n+2 + 3n - 15)}{6} = \frac{n(4n^2+9n-13)}{6}$

3. 最終的な答え

(1) $n(n+2)$

(2) $\frac{n(n+1)(n+8)}{3}$

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

(4) $n(2n^3 + 8n^2 + 11n + 6)$

(5) $\frac{n(4n^2+9n-13)}{6}$