## 1. 問題の内容

##

1. 問題の内容

次の和を$\Sigma$記号を用いて表し、和を求めよ。

(1) $1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n3^n$

(2) $1 \cdot 1 + 2 \cdot 3^{-1} + 3 \cdot 3^{-2} + \dots + n3^{1-n}$

(3) $1 \cdot 3 + 4 \cdot 3^2 + 7 \cdot 3^3 + \dots + (3n-2)3^n$

##

2. 解き方の手順

### (1)

$\Sigma$記号で表すと、

$$$$

S_n = \sum_{k=1}^{n} k \cdot 3^k

$$$$

となる。

$S_n = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n3^n$

$3S_n = 1 \cdot 3^2 + 2 \cdot 3^3 + \dots + (n-1)3^n + n3^{n+1}$

$S_n - 3S_n = 1 \cdot 3 + 1 \cdot 3^2 + 1 \cdot 3^3 + \dots + 1 \cdot 3^n - n3^{n+1}$

$-2S_n = 3(1 + 3 + 3^2 + \dots + 3^{n-1}) - n3^{n+1}$

$-2S_n = 3\frac{3^n - 1}{3 - 1} - n3^{n+1}$

$-2S_n = \frac{3}{2}(3^n - 1) - n3^{n+1}$

$-4S_n = 3(3^n - 1) - 2n3^{n+1}$

$4S_n = -3(3^n - 1) + 2n3^{n+1}$

$4S_n = -3^{n+1} + 3 + 2n3^{n+1}$

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

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

### (2)

$\Sigma$記号で表すと、

$$$$

S_n = \sum_{k=1}^{n} k \cdot 3^{1-k}

$$$$

となる。

$S_n = 1 \cdot 1 + 2 \cdot 3^{-1} + 3 \cdot 3^{-2} + \dots + n3^{1-n}$

$\frac{1}{3}S_n = 1 \cdot 3^{-1} + 2 \cdot 3^{-2} + \dots + (n-1)3^{1-n} + n3^{-n}$

$S_n - \frac{1}{3}S_n = 1 + 1 \cdot 3^{-1} + 1 \cdot 3^{-2} + \dots + 1 \cdot 3^{1-n} - n3^{-n}$

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

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

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

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

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

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

$4S_n = 9 - \frac{9+6n}{3^n}$

$S_n = \frac{9}{4} - \frac{9+6n}{4 \cdot 3^n}$

### (3)

$\Sigma$記号で表すと、

$$$$

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

$$$$

となる。

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

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

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

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

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

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

$-4S_n = 6 + 3(3^n-3) - 2(3n-2)3^{n+1}$

$-4S_n = 6 + 3^{n+1} - 9 - 6n3^{n+1} + 4 \cdot 3^{n+1}$

$-4S_n = -3 + 5 \cdot 3^{n+1} - 6n3^{n+1}$

$4S_n = 3 + (6n-5)3^{n+1}$

$S_n = \frac{3 + (6n-5)3^{n+1}}{4}$

##

3. 最終的な答え

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

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

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