問題は、以下の3つの和 $S$ を求めることです。 (1) $S = 1 \cdot 1 + 2 \cdot 5 + 3 \cdot 5^2 + 4 \cdot 5^3 + \dots + n \cdot 5^{n-1}$ (2) $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \dots + \frac{n}{3^{n-1}}$ (3) $S = 1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}$

1. 問題の内容

問題は、以下の3つの和 $S$ を求めることです。

(1) $S = 1 \cdot 1 + 2 \cdot 5 + 3 \cdot 5^2 + 4 \cdot 5^3 + \dots + n \cdot 5^{n-1}$

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

(3) $S = 1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}$

2. 解き方の手順

(1) $S = 1 \cdot 1 + 2 \cdot 5 + 3 \cdot 5^2 + 4 \cdot 5^3 + \dots + n \cdot 5^{n-1}$ について

$5S = 1 \cdot 5 + 2 \cdot 5^2 + 3 \cdot 5^3 + \dots + (n-1) \cdot 5^{n-1} + n \cdot 5^n$

$S - 5S = (1 \cdot 1 + 2 \cdot 5 + 3 \cdot 5^2 + \dots + n \cdot 5^{n-1}) - (1 \cdot 5 + 2 \cdot 5^2 + 3 \cdot 5^3 + \dots + (n-1) \cdot 5^{n-1} + n \cdot 5^n)$

$-4S = 1 + (2-1) \cdot 5 + (3-2) \cdot 5^2 + \dots + (n - (n-1)) \cdot 5^{n-1} - n \cdot 5^n$

$-4S = 1 + 5 + 5^2 + \dots + 5^{n-1} - n \cdot 5^n$

等比数列の和の公式より、$1 + 5 + 5^2 + \dots + 5^{n-1} = \frac{5^n - 1}{5 - 1} = \frac{5^n - 1}{4}$

$-4S = \frac{5^n - 1}{4} - n \cdot 5^n$

$-16S = 5^n - 1 - 4n \cdot 5^n$

$16S = 1 - 5^n + 4n \cdot 5^n = 1 + (4n-1)5^n$

$S = \frac{1 + (4n-1)5^n}{16}$

(2) $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \dots + \frac{n}{3^{n-1}}$ について

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

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

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

等比数列の和の公式より、$1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots + \frac{1}{3^{n-1}} = \frac{1 - (\frac{1}{3})^n}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{3^n}}{\frac{2}{3}} = \frac{3}{2} (1 - \frac{1}{3^n})$

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

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

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

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

(3) $S = 1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}$ について

$xS = x + 4x^2 + 7x^3 + \dots + (3n-5)x^{n-1} + (3n-2)x^n$

$S - xS = (1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}) - (x + 4x^2 + 7x^3 + \dots + (3n-5)x^{n-1} + (3n-2)x^n)$

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

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

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

$(1-x)^2 S = (1-x) + 3x(1-x^{n-1}) - (3n-2)x^n(1-x)$

$(1-x)^2 S = 1 - x + 3x - 3x^n - (3n-2)x^n + (3n-2)x^{n+1}$

$(1-x)^2 S = 1 + 2x - (3n+1)x^n + (3n-2)x^{n+1}$

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

3. 最終的な答え

(1) $S = \frac{1 + (4n-1)5^n}{16}$

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

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