## 解答

## 解答

### (1) 問題の内容

数列の和 $S = 1 \cdot 1 + 3 \cdot 2 + 5 \cdot 2^2 + \dots + (2n-1) \cdot 2^{n-1}$ を求める。

### (1) 解き方の手順

$S = \sum_{k=1}^{n} (2k-1)2^{k-1}$ と表せる。

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

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

$-S = 1 + \sum_{k=2}^{n} \left[ (2k-1) - (2k-3) \right] 2^{k-1} - (2n-1)2^n$

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

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

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

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

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

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

$-S = 2^{n+1} - 3 - 2n2^n + 2^n$

$-S = 2^{n+1} + 2^n - 2n2^n - 3$

$-S = 3 \cdot 2^n - 2n2^n - 3$

$S = (2n-3)2^n + 3$

### (1) 最終的な答え

$S = (2n-3)2^n + 3$

### (2) 問題の内容

数列の和 $S = 1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}$ を求める。

### (2) 解き方の手順

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

$xS = \sum_{k=1}^n (3k-2)x^k = \sum_{k=2}^{n+1} (3k-5)x^{k-1}$

$S - xS = \sum_{k=1}^n (3k-2)x^{k-1} - \sum_{k=2}^{n+1} (3k-5)x^{k-1}$

$(1-x)S = 1 + \sum_{k=2}^n \left[ (3k-2) - (3k-5) \right] x^{k-1} - (3n-2)x^n$

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

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

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

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

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

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

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

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

### (2) 最終的な答え

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

### (3) 問題の内容

数列の和 $S = 2^{n-1} + 2 \cdot 2^{n-2} + 3 \cdot 2^{n-3} + \dots + (n-1) \cdot 2 + n$ を求める。

### (3) 解き方の手順

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

$S = 2^{n-1} + 2 \cdot 2^{n-2} + 3 \cdot 2^{n-3} + \dots + (n-1) \cdot 2^1 + n \cdot 2^0$

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

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

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

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

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

$S = 2^{n+1} - 2 - n$

### (3) 最終的な答え

$S = 2^{n+1} - n - 2$