## 問題230の解答

## 問題230の解答

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1. 問題の内容

与えられた6つの和の計算問題を解く。

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

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2. 解き方の手順

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

和を分解し、$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ を利用する。

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

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

和を分解し、$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ と $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ を利用する。

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

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

和を分解し、$\sum_{k=1}^{n} k^3 = (\frac{n(n+1)}{2})^2$ を利用する。

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

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

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

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

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

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

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

$\sum_{k=1}^{n} 3^k$ は初項3、公比3の等比数列の和であるから、 $\sum_{k=1}^{n} 3^k = \frac{3(3^n - 1)}{3 - 1} = \frac{3(3^n - 1)}{2}$

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

したがって、 $\sum_{k=1}^{n} (3^k + 2) = \frac{3(3^n - 1)}{2} + 2n = \frac{3^{n+1} - 3}{2} + 2n = \frac{3^{n+1} - 3 + 4n}{2}$

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