与えられた4つの数列の和をそれぞれ計算します。 (1) $\sum_{k=1}^{n} (4k+3)$ (2) $\sum_{k=1}^{n} (3k^2 - 7k + 4)$ (3) $\sum_{k=1}^{n} k(k+2)$ (4) $\sum_{k=1}^{n-1} 5k$

1. 問題の内容

与えられた4つの数列の和をそれぞれ計算します。

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

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

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

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

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

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

(3) $\sum_{k=1}^{n} k(k+2) = \sum_{k=1}^{n} (k^2 + 2k) = \sum_{k=1}^{n} k^2 + 2\sum_{k=1}^{n} k = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{6} + n(n+1) = \frac{n(n+1)(2n+1) + 6n(n+1)}{6} = \frac{n(n+1)(2n+1 + 6)}{6} = \frac{n(n+1)(2n+7)}{6} = \frac{n(2n^2 + 9n + 7)}{6} = \frac{2n^3 + 9n^2 + 7n}{6}$

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

3. 最終的な答え

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

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

(3) $\frac{2n^3 + 9n^2 + 7n}{6}$

(4) $\frac{5n^2 - 5n}{2}$