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

1. 問題の内容

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

2. 解き方の手順

まず、数列 $\{a_n\}$ の一般項を求めます。初項が3、公比が $\frac{1}{5}$ の等比数列なので、

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

となります。

次に、 $S_n$ の式に $a_k$ を代入します。

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

ここで、$\sum_{k=1}^{n} k^2 (\frac{1}{5})^{k-1}$ の部分を計算します。

$T_n = \sum_{k=1}^{n} k^2 (\frac{1}{5})^{k-1} = 1^2 + 2^2 (\frac{1}{5}) + 3^2 (\frac{1}{5})^2 + \cdots + n^2 (\frac{1}{5})^{n-1}$

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

$T_n - \frac{1}{5} T_n = \frac{4}{5} T_n = 1 + (2^2 - 1^2) (\frac{1}{5}) + (3^2 - 2^2) (\frac{1}{5})^2 + \cdots + (n^2 - (n-1)^2) (\frac{1}{5})^{n-1} - n^2 (\frac{1}{5})^n$

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

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

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

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

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

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

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

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

$U = 2 ( \frac{5}{16} (1 - \frac{1}{5^{n-1}}) - \frac{5(n-1)}{4 \cdot 5^n} ) + \frac{1-(\frac{1}{5})^{n-1}}{4} = \frac{5}{8} (1 - \frac{1}{5^{n-1}}) - \frac{5(n-1)}{2 \cdot 5^n} + \frac{1}{4} (1 - \frac{1}{5^{n-1}})$

$= \frac{7}{8} - \frac{7}{8 \cdot 5^{n-1}} - \frac{5(n-1)}{2 \cdot 5^n} = \frac{7}{8} - \frac{35}{8 \cdot 5^n} - \frac{10(n-1)}{4 \cdot 5^n}$

$\frac{4}{5} T_n = 1 + (\frac{7}{8} - \frac{35}{8 \cdot 5^n} - \frac{5(n-1)}{2 \cdot 5^n}) - \frac{n^2}{5^n} = \frac{15}{8} - (\frac{35}{8} + \frac{20(n-1)}{8} + \frac{8 n^2}{8} ) \frac{1}{5^n}$

$\frac{4}{5} T_n = \frac{15}{8} - \frac{8n^2 + 20n - 20 + 35}{8 \cdot 5^n} = \frac{15}{8} - \frac{8n^2 + 20n + 15}{8 \cdot 5^n}$

$T_n = \frac{5}{4} (\frac{15}{8} - \frac{8n^2 + 20n + 15}{8 \cdot 5^n}) = \frac{75}{32} - \frac{5(8n^2 + 20n + 15)}{32 \cdot 5^n}$

$S_n = 3 T_n = 3 (\frac{75}{32} - \frac{5(8n^2 + 20n + 15)}{32 \cdot 5^n}) = \frac{225}{32} - \frac{15(8n^2 + 20n + 15)}{32 \cdot 5^n}$

3. 最終的な答え

$S_n = \frac{225}{32} - \frac{15(8n^2 + 20n + 15)}{32 \cdot 5^n}$