以下の6つの数列の和を計算します。 (1) $\sum_{k=1}^{n} (2k-7)$ (2) $\sum_{k=1}^{n} (5k^2-4k+2)$ (3) $\sum_{k=1}^{n} (4k^3-1)$ (4) $\sum_{k=1}^{n} k(k+1)$ (5) $\sum_{m=1}^{n} (3m-1)^2$ (6) $\sum_{k=1}^{n} (3^k + 2)$

はい、承知いたしました。画像に写っている問題のうち、問題(1)〜(6)を解きます。

1. 問題の内容

以下の6つの数列の和を計算します。

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

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

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

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

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

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

2. 解き方の手順

数列の和の公式を利用して計算します。

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

(2) $\sum_{k=1}^{n} (5k^2-4k+2) = 5 \sum_{k=1}^{n} k^2 - 4 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 2 = 5 \cdot \frac{n(n+1)(2n+1)}{6} - 4 \cdot \frac{n(n+1)}{2} + 2n = \frac{5n(n+1)(2n+1)}{6} - 2n(n+1) + 2n = \frac{5n(2n^2+3n+1)}{6} - 2n^2 - 2n + 2n = \frac{10n^3+15n^2+5n}{6} - 2n^2 = \frac{10n^3+15n^2+5n-12n^2}{6} = \frac{10n^3+3n^2+5n}{6}$

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

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

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

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

3. 最終的な答え

(1) $n^2 - 6n$

(2) $\frac{10n^3+3n^2+5n}{6}$

(3) $n^4+2n^3+n^2-n$

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

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

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