次の6つの和を計算します。 (1) $\sum_{k=1}^{30} k$ (2) $\sum_{k=1}^{n} (3k-2)$ (3) $\sum_{k=1}^{n} 3^k$ (4) $\sum_{k=1}^{n} \frac{1}{2^{k-1}}$ (5) $\sum_{k=0}^{n-1} (k+1)(k-1)$ (6) $\sum_{k=1}^{n-1} (2 \cdot 3^k - 4^{k-1})$

1. 問題の内容

次の6つの和を計算します。

(1) $\sum_{k=1}^{30} k$

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

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

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

(5) $\sum_{k=0}^{n-1} (k+1)(k-1)$

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

2. 解き方の手順

(1) 等差数列の和の公式 $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ を用います。

$n = 30$ を代入すると、$\sum_{k=1}^{30} k = \frac{30(30+1)}{2} = \frac{30 \cdot 31}{2} = 15 \cdot 31 = 465$

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

(3) 等比数列の和の公式 $\sum_{k=1}^{n} ar^{k-1} = a \cdot \frac{1-r^n}{1-r}$ を用います。

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

(4) 等比数列の和の公式 $\sum_{k=1}^{n} ar^{k-1} = a \cdot \frac{1-r^n}{1-r}$ を用います。

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

(5) $\sum_{k=0}^{n-1} (k+1)(k-1) = \sum_{k=0}^{n-1} (k^2 - 1) = \sum_{k=0}^{n-1} k^2 - \sum_{k=0}^{n-1} 1 = \sum_{k=1}^{n-1} k^2 - n = \frac{(n-1)n(2n-1)}{6} - n = \frac{(n-1)n(2n-1) - 6n}{6} = \frac{n((n-1)(2n-1) - 6)}{6} = \frac{n(2n^2 - 3n + 1 - 6)}{6} = \frac{n(2n^2 - 3n - 5)}{6} = \frac{n(n+1)(2n-5)}{6}$

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

3. 最終的な答え

(1) 465

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

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

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

(5) $\frac{n(n+1)(2n-5)}{6}$

(6) $3^n - \frac{8}{3} - \frac{4^{n-1}}{3}$