数列 $\{a_n\}$ は、初項が 3、公比が $\frac{1}{5}$ の等比数列である。このとき、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。このとき、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。初項が 3、公比が $\frac{1}{5}$ の等比数列なので、

$a_n = 3 \cdot \left(\frac{1}{5}\right)^{n-1}$

である。

次に、$S_n$ の式に $a_k$ を代入する。

$S_n = \sum_{k=1}^{n} k^2 a_k = \sum_{k=1}^{n} k^2 \cdot 3 \cdot \left(\frac{1}{5}\right)^{k-1} = 3 \sum_{k=1}^{n} k^2 \cdot \left(\frac{1}{5}\right)^{k-1}$

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

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

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

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

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

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

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

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

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

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

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

$\frac{4}{5} V_n = \frac{1}{4} (1- (\frac{1}{5})^{n-1} ) - (n-1)(\frac{1}{5})^n$

$V_n = \frac{5}{16} (1- (\frac{1}{5})^{n-1} ) - \frac{5(n-1)}{4} (\frac{1}{5})^n = \frac{5}{16} (1- (\frac{1}{5})^{n-1} ) - \frac{(n-1)}{4 \cdot 5^{n-1}}$

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

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

$\frac{4}{5} T_n = \frac{13}{8} - \frac{5}{8} (\frac{1}{5})^{n-1} - \frac{(n-1)}{2} (\frac{1}{5})^{n-1} (\frac{1}{5}) - n^2 (\frac{1}{5})^n - \frac{1}{4 \cdot 5^{n-1}}$

3. 最終的な答え

$S_n = \frac{125}{32} - \frac{15(4n^2+10n+21)}{32 \cdot 5^n}$