次の6つの和の計算問題を解きます。 (1) $\sum_{k=1}^{n} (2k+3)$ (2) $\sum_{k=1}^{n} (k^2+k)$ (3) $\sum_{k=1}^{n} (k^2-6k+5)$ (4) $\sum_{k=1}^{n} (k^3-4k)$ (5) $\sum_{k=1}^{n} (k+1)(k-2)$ (6) $\sum_{k=1}^{n-1} (k^2-5k)$

1. 問題の内容

次の6つの和の計算問題を解きます。

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

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

(3) $\sum_{k=1}^{n} (k^2-6k+5)$

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

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

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

2. 解き方の手順

これらの和を計算するために、次の公式を使用します。

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

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

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

(1)

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

(2)

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

(3)

$\sum_{k=1}^{n} (k^2-6k+5) = \sum_{k=1}^{n} k^2 - 6\sum_{k=1}^{n} k + 5\sum_{k=1}^{n} 1 = \frac{n(n+1)(2n+1)}{6} - 6\frac{n(n+1)}{2} + 5n = \frac{n(n+1)(2n+1)}{6} - 3n(n+1) + 5n = \frac{n(n+1)(2n+1) - 18n(n+1) + 30n}{6} = \frac{n(2n^2+3n+1 - 18n - 18 + 30)}{6} = \frac{n(2n^2-15n+13)}{6}$

(4)

$\sum_{k=1}^{n} (k^3-4k) = \sum_{k=1}^{n} k^3 - 4\sum_{k=1}^{n} k = \left(\frac{n(n+1)}{2}\right)^2 - 4\frac{n(n+1)}{2} = \frac{n^2(n+1)^2}{4} - 2n(n+1) = \frac{n^2(n+1)^2 - 8n(n+1)}{4} = \frac{n(n+1)(n(n+1)-8)}{4} = \frac{n(n+1)(n^2+n-8)}{4}$

(5)

$\sum_{k=1}^{n} (k+1)(k-2) = \sum_{k=1}^{n} (k^2 - 2k + k - 2) = \sum_{k=1}^{n} (k^2 - k - 2) = \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k - 2\sum_{k=1}^{n} 1 = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} - 2n = \frac{n(n+1)(2n+1) - 3n(n+1) - 12n}{6} = \frac{n(2n^2+3n+1 - 3n - 3 - 12)}{6} = \frac{n(2n^2-14)}{6} = \frac{2n(n^2-7)}{6} = \frac{n(n^2-7)}{3}$

(6)

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

3. 最終的な答え

(1) $n(n+4)$

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

(3) $\frac{n(2n^2-15n+13)}{6}$

(4) $\frac{n(n+1)(n^2+n-8)}{4}$

(5) $\frac{n(n^2-7)}{3}$

(6) $\frac{n(n-1)(n-8)}{3}$