数列 $\{a_n\}$ が漸化式 $a_1 = 1$, $a_{n+1} = 2a_n + n^2$ ($n=1, 2, 3, \dots$) で定義されているとき、一般項 $a_n$ を求める問題です。

1. 問題の内容

数列 $\{a_n\}$ が漸化式 $a_1 = 1$, $a_{n+1} = 2a_n + n^2$ ($n=1, 2, 3, \dots$) で定義されているとき、一般項 $a_n$ を求める問題です。

2. 解き方の手順

まず、$a_{n+1} = 2a_n + n^2$ の両辺を $2^{n+1}$ で割ります。

$\frac{a_{n+1}}{2^{n+1}} = \frac{a_n}{2^n} + \frac{n^2}{2^{n+1}}$

ここで、$b_n = \frac{a_n}{2^n}$ とおくと、

$b_{n+1} = b_n + \frac{n^2}{2^{n+1}}$

$b_{n+1} - b_n = \frac{n^2}{2^{n+1}}$

$n \ge 2$ のとき、

$b_n = b_1 + \sum_{k=1}^{n-1} (b_{k+1} - b_k) = b_1 + \sum_{k=1}^{n-1} \frac{k^2}{2^{k+1}}$

$b_1 = \frac{a_1}{2^1} = \frac{1}{2}$

したがって、

$b_n = \frac{1}{2} + \sum_{k=1}^{n-1} \frac{k^2}{2^{k+1}}$

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

ここで、$S = \sum_{k=1}^{n-1} k^2 x^k$ を計算します ($x = \frac{1}{2}$)。

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

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

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

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

$\sum_{k=1}^{n-1} x^k = \frac{x(1-x^{n-1})}{1-x}$

$\sum_{k=1}^{n-1} kx^k = x \frac{d}{dx} \sum_{k=1}^{n-1} x^k = x \frac{d}{dx} \frac{x(1-x^{n-1})}{1-x} = x \frac{(1-(n-1)x^{n-2})(1-x) - x(1-x^{n-1})(-1)}{(1-x)^2}$

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

$x = \frac{1}{2}$ のとき、

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

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

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

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

$S = 6 - \frac{n^2+2n+1}{2^{n-1}}$

$b_n = \frac{1}{2} + \frac{1}{2} \sum_{k=1}^{n-1} \frac{k^2}{2^k} = \frac{1}{2} + \frac{1}{2} (6 - \frac{n^2 + 2n + 1}{2^{n-1}}) = \frac{7}{2} - \frac{n^2+2n+1}{2^n}$

$a_n = 2^n b_n = 2^n (\frac{7}{2} - \frac{n^2+2n+1}{2^n}) = 7 \cdot 2^{n-1} - (n^2+2n+1)$

$a_n = 7 \cdot 2^{n-1} - (n+1)^2$

3. 最終的な答え

$a_n = 7 \cdot 2^{n-1} - (n+1)^2$