与えられた数列の和を $\Sigma$ 記号を用いて表し、その和を求める問題です。 (1) $1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n 3^n$ (2) $1 \cdot 1 + 2 \cdot 3^{-1} + 3 \cdot 3^{-2} + \dots + n 3^{1-n}$ (3) $1 \cdot 3 + 4 \cdot 3^2 + 7 \cdot 3^3 + \dots + (3n-2) 3^n$

1. 問題の内容

与えられた数列の和を $\Sigma$ 記号を用いて表し、その和を求める問題です。

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

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

(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) の場合：

$\Sigma$ 記号を用いた表現：

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

和の計算：

$S = \sum_{k=1}^{n} k \cdot 3^k = 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n$

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

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

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

等比数列の和の公式 $\sum_{k=1}^{n} ar^{k-1} = a \frac{1-r^n}{1-r}$ より、

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

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

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

$S = \frac{1}{4} [2n \cdot 3^{n+1} - 3^{n+1} + 3]$

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

(2) の場合：

$\Sigma$ 記号を用いた表現：

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

和の計算：

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

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

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

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

等比数列の和の公式 $\sum_{k=1}^{n} ar^{k-1} = a \frac{1-r^n}{1-r}$ より、

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

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

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

$S = \frac{27}{4} - \frac{27}{4 \cdot 3^n} - \frac{3n}{2 \cdot 3^n}$

$S = \frac{27}{4} - \frac{27 + 6n}{4 \cdot 3^n} = \frac{27 \cdot 3^n - 27 - 6n}{4 \cdot 3^n} = \frac{27 (3^n - 1) - 6n}{4 \cdot 3^n}$

(3) の場合：

$\Sigma$ 記号を用いた表現：

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

和の計算：

$S = \sum_{k=1}^{n} (3k-2) \cdot 3^k = 1 \cdot 3 + 4 \cdot 3^2 + 7 \cdot 3^3 + \dots + (3n-2) \cdot 3^n$

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

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

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

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

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

$-2S = 3 - \frac{27}{2} (1-3^{n-1}) - (3n-2) \cdot 3^{n+1}$

$-2S = 3 - \frac{27}{2} + \frac{27}{2} \cdot 3^{n-1} - (3n-2) \cdot 3^{n+1}$

$-2S = -\frac{21}{2} + \frac{9}{2} \cdot 3^n - (9n-6) \cdot 3^n$

$-2S = -\frac{21}{2} + (\frac{9}{2} - 9n + 6) 3^n$

$-2S = -\frac{21}{2} + (\frac{21}{2} - 9n) 3^n$

$S = \frac{21}{4} - (\frac{21}{4} - \frac{9}{2} n) 3^n$

$S = \frac{21}{4} - \frac{3}{4} (7 - 6n) 3^n = \frac{3}{4} [7 - (7 - 6n) 3^n]$

3. 最終的な答え

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

(2) $\sum_{k=1}^{n} k \cdot 3^{1-k} = \frac{27 (3^n - 1) - 6n}{4 \cdot 3^n}$

(3) $\sum_{k=1}^{n} (3k-2) \cdot 3^k = \frac{3}{4} [7 - (7 - 6n) 3^n]$