与えられた数列の和 $S_n$ を求める問題です。 (1) $S_n = 1 \cdot 1 + 2 \cdot 3 + 3 \cdot 3^2 + 4 \cdot 3^3 + \cdots + n \cdot 3^{n-1}$ (2) $S_n = 1 \cdot r + 3 \cdot r^2 + 5 \cdot r^3 + 7 \cdot r^4 + \cdots + (2n-1) \cdot r^n$ ($r \neq 1$)

1. 問題の内容

与えられた数列の和 $S_n$ を求める問題です。

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

(2) $S_n = 1 \cdot r + 3 \cdot r^2 + 5 \cdot r^3 + 7 \cdot r^4 + \cdots + (2n-1) \cdot r^n$ ($r \neq 1$)

2. 解き方の手順

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

$3S_n = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \cdots + (n-1) \cdot 3^{n-1} + n \cdot 3^n$

$S_n - 3S_n = (1 \cdot 1 + 2 \cdot 3 + 3 \cdot 3^2 + \cdots + n \cdot 3^{n-1}) - (1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \cdots + (n-1) \cdot 3^{n-1} + n \cdot 3^n)$

$-2S_n = 1 + 3 + 3^2 + 3^3 + \cdots + 3^{n-1} - n \cdot 3^n$

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

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

(2) $S_n = 1 \cdot r + 3 \cdot r^2 + 5 \cdot r^3 + 7 \cdot r^4 + \cdots + (2n-1) \cdot r^n$

$rS_n = 1 \cdot r^2 + 3 \cdot r^3 + 5 \cdot r^4 + \cdots + (2n-3) \cdot r^n + (2n-1) \cdot r^{n+1}$

$S_n - rS_n = (1 \cdot r + 3 \cdot r^2 + 5 \cdot r^3 + \cdots + (2n-1) \cdot r^n) - (1 \cdot r^2 + 3 \cdot r^3 + 5 \cdot r^4 + \cdots + (2n-3) \cdot r^n + (2n-1) \cdot r^{n+1})$

$(1-r)S_n = r + 2r^2 + 2r^3 + 2r^4 + \cdots + 2r^n - (2n-1)r^{n+1}$

$(1-r)S_n = r + 2(r^2 + r^3 + \cdots + r^n) - (2n-1)r^{n+1}$

$(1-r)S_n = r + 2\frac{r^2(r^{n-1}-1)}{r-1} - (2n-1)r^{n+1} = r + \frac{2r^2(r^{n-1}-1)}{r-1} - (2n-1)r^{n+1}$

$(1-r)S_n = r + \frac{2r^{n+1} - 2r^2}{r-1} - (2n-1)r^{n+1}$

$(1-r)S_n = \frac{r(r-1) + 2r^{n+1} - 2r^2 - (2n-1)r^{n+1}(r-1)}{r-1} = \frac{r^2-r+2r^{n+1}-2r^2 - (2n-1)r^{n+2} + (2n-1)r^{n+1}}{r-1}$

$(1-r)S_n = \frac{-r^2-r + (2n)r^{n+1} - (2n-1)r^{n+2} }{r-1}$

$S_n = \frac{-r^2-r+(2n)r^{n+1} -(2n-1)r^{n+2}}{(r-1)(1-r)}$

$S_n = \frac{r(1+r) - (2n)r^{n+1} + (2n-1)r^{n+2}}{(1-r)^2}$

3. 最終的な答え

(1) $S_n = \frac{(1-2n)3^n - 1}{4}$

(2) $S_n = \frac{r(1+r) - (2n)r^{n+1} + (2n-1)r^{n+2}}{(1-r)^2}$