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 $\{n^2 a_n\}$ の初項から第n項までの和 $S_n = \sum_{k=1}^n k^2 a_k$ を求めよ。

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 $\{n^2 a_n\}$ の初項から第n項までの和 $S_n = \sum_{k=1}^n k^2 a_k$ を求めよ。

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$\frac{1}{5}$なので、

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

したがって、

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

$T_n = \sum_{k=1}^n k^2 x^{k-1}$ とおくと、$S_n = 3T_n$ （ただし、$x=\frac{1}{5}$）。

$$T_n = 1^2 + 2^2x + 3^2x^2 + \cdots + n^2 x^{n-1}$$

両辺に $x$ をかけると、

$$xT_n = 1^2x + 2^2x^2 + 3^2x^3 + \cdots + n^2 x^{n}$$

$T_n - xT_n = (1-x)T_n$ を計算する。

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

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

$U_n = 1 + 3x + 5x^2 + \cdots + (2n-1)x^{n-1}$ とおくと、

$$xU_n = x + 3x^2 + 5x^3 + \cdots + (2n-1)x^n$$

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

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

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

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

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

$$(1-x)T_n = U_n - n^2x^n$$

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

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

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

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

$x = \frac{1}{5}$ なので、$1-x = \frac{4}{5}$, $(1-x)^2 = \frac{16}{25}$, $(1-x)^3 = \frac{64}{125}$

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

$$T_n = \frac{\frac{6}{5} - (2n+1+n^2) (\frac{1}{5})^n + (2n-1+2n^2) (\frac{1}{5})^{n+1} - n^2 (\frac{1}{5})^{n+2}}{\frac{64}{125}}$$

$$T_n = \frac{125}{64} \left[ \frac{6}{5} - \left(2n+1+n^2\right) \left(\frac{1}{5}\right)^n + \left(2n-1+2n^2\right) \left(\frac{1}{5}\right)^{n+1} - n^2 \left(\frac{1}{5}\right)^{n+2} \right]$$

$$S_n = 3 T_n = \frac{375}{64} \left[ \frac{6}{5} - \left(2n+1+n^2\right) \left(\frac{1}{5}\right)^n + \left(2n-1+2n^2\right) \left(\frac{1}{5}\right)^{n+1} - n^2 \left(\frac{1}{5}\right)^{n+2} \right]$$

3. 最終的な答え

$$S_n = \frac{375}{64} \left[ \frac{6}{5} - \frac{2n+1+n^2}{5^n} + \frac{2n-1+2n^2}{5^{n+1}} - \frac{n^2}{5^{n+2}} \right]$$

$$S_n = \frac{450}{64} - \frac{375}{64} \left[ \frac{2n+1+n^2}{5^n} - \frac{2n-1+2n^2}{5^{n+1}} + \frac{n^2}{5^{n+2}} \right]$$

$$S_n = \frac{225}{32} - \frac{375}{64 \cdot 5^{n+2}} [25(n^2+2n+1) - 5(2n^2+2n-1)+n^2]$$

$$S_n = \frac{225}{32} - \frac{375}{64 \cdot 5^{n+2}} [25n^2+50n+25 - 10n^2 - 10n+5 + n^2]$$

$$S_n = \frac{225}{32} - \frac{375}{64 \cdot 5^{n+2}} [16n^2 + 40n + 30]$$

$$S_n = \frac{225}{32} - \frac{375}{64 \cdot 5^{n+2}} [2(8n^2 + 20n + 15)] = \frac{225}{32} - \frac{375}{32 \cdot 5^{n+2}} [\frac{(8n^2+20n+15)}{5}]$$

$$S_n = \frac{225}{32} - \frac{750(8n^2+20n+15)}{64(5^{n+2})}$$

$$S_n = \frac{225}{32} - \frac{375(8n^2+20n+15)}{32 \cdot 5^{n+3}}$$

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

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