次の和 $S$ を求めます。 (1) $S = 1 \cdot 1 + 2 \cdot 4 + 3 \cdot 4^2 + \dots + n \cdot 4^{n-1}$ (2) $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \dots + \frac{n}{3^{n-1}}$

1. 問題の内容

次の和 $S$ を求めます。

(1) $S = 1 \cdot 1 + 2 \cdot 4 + 3 \cdot 4^2 + \dots + n \cdot 4^{n-1}$

(2) $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \dots + \frac{n}{3^{n-1}}$

2. 解き方の手順

(1) $S = 1 \cdot 1 + 2 \cdot 4 + 3 \cdot 4^2 + \dots + n \cdot 4^{n-1}$

両辺を4倍すると、

$4S = 1 \cdot 4 + 2 \cdot 4^2 + 3 \cdot 4^3 + \dots + (n-1) \cdot 4^{n-1} + n \cdot 4^n$

$S - 4S = (1 \cdot 1 + 2 \cdot 4 + 3 \cdot 4^2 + \dots + n \cdot 4^{n-1}) - (1 \cdot 4 + 2 \cdot 4^2 + 3 \cdot 4^3 + \dots + (n-1) \cdot 4^{n-1} + n \cdot 4^n)$

$-3S = 1 + 4 + 4^2 + \dots + 4^{n-1} - n \cdot 4^n$

$1 + 4 + 4^2 + \dots + 4^{n-1}$ は初項1, 公比4, 項数nの等比数列の和なので、$\frac{1(4^n - 1)}{4 - 1} = \frac{4^n - 1}{3}$

$-3S = \frac{4^n - 1}{3} - n \cdot 4^n$

$-3S = \frac{4^n - 1 - 3n \cdot 4^n}{3}$

$S = \frac{3n \cdot 4^n - 4^n + 1}{9} = \frac{(3n-1)4^n + 1}{9}$

(2) $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \dots + \frac{n}{3^{n-1}}$

両辺を $\frac{1}{3}$ 倍すると、

$\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \dots + \frac{n-1}{3^{n-1}} + \frac{n}{3^n}$

$S - \frac{1}{3}S = (1 + \frac{2}{3} + \frac{3}{3^2} + \dots + \frac{n}{3^{n-1}}) - (\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \dots + \frac{n-1}{3^{n-1}} + \frac{n}{3^n})$

$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^{n-1}} - \frac{n}{3^n}$

$1 + \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^{n-1}}$ は初項1, 公比 $\frac{1}{3}$, 項数nの等比数列の和なので、$\frac{1(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} = \frac{1 - (\frac{1}{3})^n}{\frac{2}{3}} = \frac{3}{2}(1 - \frac{1}{3^n}) = \frac{3}{2} - \frac{3}{2} \cdot \frac{1}{3^n} = \frac{3}{2} - \frac{1}{2 \cdot 3^{n-1}}$

$\frac{2}{3}S = \frac{3}{2} - \frac{1}{2 \cdot 3^{n-1}} - \frac{n}{3^n} = \frac{3}{2} - \frac{3}{2 \cdot 3^n} - \frac{2n}{2 \cdot 3^n} = \frac{3}{2} - \frac{3 + 2n}{2 \cdot 3^n}$

$S = \frac{3}{2} \cdot \frac{3}{2} - \frac{3 + 2n}{2 \cdot 3^n} \cdot \frac{3}{2} = \frac{9}{4} - \frac{3(3 + 2n)}{4 \cdot 3^n} = \frac{9}{4} - \frac{3 + 2n}{4 \cdot 3^{n-1}} = \frac{9 \cdot 3^{n-1} - 3 - 2n}{4 \cdot 3^{n-1}}$

$S = \frac{9 \cdot 3^{n-1} - 2n - 3}{4 \cdot 3^{n-1}}$

3. 最終的な答え

(1) $S = \frac{(3n-1)4^n + 1}{9}$

(2) $S = \frac{9 \cdot 3^{n-1} - 2n - 3}{4 \cdot 3^{n-1}}$