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

1. 問題の内容

次の3つの和を求める問題です。

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

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

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

2. 解き方の手順

(1)

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

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

$\sum_{k=1}^{n} 3 = 3n$

よって、

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

(2)

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

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

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

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

よって、

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

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

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

$= n^3 + 2n^2 - n$

(3)

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

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

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

よって、

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

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

$= \frac{n^4 + 2n^3 + 2n^2 + n}{2}$

3. 最終的な答え

(1) $n(2n+5)$

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

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