数列 $\{a_n\}$ は初項が3、公比が $\frac{1}{5}$ の等比数列である。数列 $\{n^2a_n\}$ の初項から第 $n$ 項までの和 $S_n = \sum_{k=1}^n k^2 a_k$ を求める。

1. 問題の内容

数列 $\{a_n\}$ は初項が3、公比が $\frac{1}{5}$ の等比数列である。数列 $\{n^2a_n\}$ の初項から第 $n$ 項までの和 $S_n = \sum_{k=1}^n k^2 a_k$ を求める。

2. 解き方の手順

まず、数列 $\{a_n\}$ の一般項を求める。等比数列の一般項は $a_n = a_1 r^{n-1}$ で表されるので、$a_n = 3 \cdot \left(\frac{1}{5}\right)^{n-1}$ である。

次に、$S_n$ の式に代入して $S_n = \sum_{k=1}^n k^2 \cdot 3 \cdot \left(\frac{1}{5}\right)^{k-1} = 3 \sum_{k=1}^n k^2 \left(\frac{1}{5}\right)^{k-1}$ となる。

$T_n = \sum_{k=1}^n k^2 \left(\frac{1}{5}\right)^{k-1}$ とおくと、

$T_n = 1^2 + 2^2 \cdot \frac{1}{5} + 3^2 \cdot \left(\frac{1}{5}\right)^2 + \cdots + n^2 \left(\frac{1}{5}\right)^{n-1}$

$\frac{1}{5} T_n = 1^2 \cdot \frac{1}{5} + 2^2 \cdot \left(\frac{1}{5}\right)^2 + \cdots + (n-1)^2 \left(\frac{1}{5}\right)^{n-1} + n^2 \left(\frac{1}{5}\right)^n$

$T_n - \frac{1}{5} T_n = 1 + \sum_{k=2}^n (k^2 - (k-1)^2) \left(\frac{1}{5}\right)^{k-1} - n^2 \left(\frac{1}{5}\right)^n$

$\frac{4}{5} T_n = 1 + \sum_{k=2}^n (2k-1) \left(\frac{1}{5}\right)^{k-1} - n^2 \left(\frac{1}{5}\right)^n$

$\frac{4}{5} T_n = 1 + \sum_{k=1}^{n-1} (2(k+1)-1) \left(\frac{1}{5}\right)^k - n^2 \left(\frac{1}{5}\right)^n$

$\frac{4}{5} T_n = 1 + \sum_{k=1}^{n-1} (2k+1) \left(\frac{1}{5}\right)^k - n^2 \left(\frac{1}{5}\right)^n$

$U_{n-1} = \sum_{k=1}^{n-1} (2k+1) \left(\frac{1}{5}\right)^k$ とおくと、

$U_{n-1} = 3 \cdot \frac{1}{5} + 5 \cdot \left(\frac{1}{5}\right)^2 + \cdots + (2(n-1)+1) \left(\frac{1}{5}\right)^{n-1}$

$\frac{1}{5} U_{n-1} = 3 \cdot \left(\frac{1}{5}\right)^2 + 5 \cdot \left(\frac{1}{5}\right)^3 + \cdots + (2(n-2)+1) \left(\frac{1}{5}\right)^{n-1} + (2(n-1)+1) \left(\frac{1}{5}\right)^n$

$U_{n-1} - \frac{1}{5} U_{n-1} = \frac{3}{5} + \sum_{k=2}^{n-1} 2 \left(\frac{1}{5}\right)^k - (2n-1) \left(\frac{1}{5}\right)^n$

$\frac{4}{5} U_{n-1} = \frac{3}{5} + 2 \sum_{k=2}^{n-1} \left(\frac{1}{5}\right)^k - (2n-1) \left(\frac{1}{5}\right)^n$

$\frac{4}{5} U_{n-1} = \frac{3}{5} + 2 \left(\frac{\frac{1}{25} \left(1 - \left(\frac{1}{5}\right)^{n-2} \right)}{1 - \frac{1}{5}}\right) - (2n-1) \left(\frac{1}{5}\right)^n$

$\frac{4}{5} U_{n-1} = \frac{3}{5} + \frac{1}{2} \left(\frac{1}{5} - \left(\frac{1}{5}\right)^{n-1} \right) - (2n-1) \left(\frac{1}{5}\right)^n$

$\frac{4}{5} U_{n-1} = \frac{3}{5} + \frac{1}{10} - \frac{1}{2} \left(\frac{1}{5}\right)^{n-1} - (2n-1) \left(\frac{1}{5}\right)^n = \frac{7}{10} - \frac{5}{2} \left(\frac{1}{5}\right)^n - (2n-1)\left(\frac{1}{5}\right)^n$

$U_{n-1} = \frac{35}{8} - \frac{25}{8} \left(\frac{1}{5}\right)^n - \frac{5}{4} (2n-1) \left(\frac{1}{5}\right)^n = \frac{35}{8} - \frac{5}{8}(10n-5+5)(\frac{1}{5})^n = \frac{35}{8} - \frac{50n}{8} (\frac{1}{5})^n = \frac{35}{8} - \frac{25}{4} (\frac{1}{5})^{n-1} \frac{n}{5}$

$\frac{4}{5} T_n = 1 + U_{n-1} - n^2 \left(\frac{1}{5}\right)^n = 1 + \frac{35}{8} - \frac{5n}{4}(\frac{1}{5})^n - n^2 \left(\frac{1}{5}\right)^n = \frac{43}{8} - \frac{5n + 4n^2}{4} \left(\frac{1}{5}\right)^n$

$T_n = \frac{5}{4} \left(\frac{43}{8} - \frac{5n + 4n^2}{4} \left(\frac{1}{5}\right)^n \right) = \frac{215}{32} - \frac{5n + 4n^2}{16} \left(\frac{1}{5}\right)^{n-1}$

$S_n = 3 T_n = \frac{645}{32} - \frac{3(5n+4n^2)}{16}(\frac{1}{5})^{n-1}$

3. 最終的な答え

$S_n = \frac{645}{32} - \frac{3(4n^2+5n)}{16} \cdot \left(\frac{1}{5}\right)^{n-1}$

または

$S_n = \frac{645}{32} - \frac{12n^2+15n}{16} \cdot \left(\frac{1}{5}\right)^{n-1}$