問題は、数列 $S$ の和を求める問題です。 $S = 1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}$

1. 問題の内容

問題は、数列 $S$ の和を求める問題です。

$S = 1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}$

2. 解き方の手順

数列の和を求めるために、一般的な方法として、$xS$ を計算し、$S$ から $xS$ を引くことを試みます。

$S = 1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}$

$xS = x + 4x^2 + 7x^3 + 10x^4 + \dots + (3n-2)x^{n}$

$S - xS = (1 + 4x + 7x^2 + 10x^3 + \dots + (3n-2)x^{n-1}) - (x + 4x^2 + 7x^3 + 10x^4 + \dots + (3n-2)x^{n})$

$S - xS = 1 + (4x - x) + (7x^2 - 4x^2) + (10x^3 - 7x^3) + \dots + ((3n-2)x^{n-1} - (3n-5)x^{n-1}) - (3n-2)x^n$

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

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

$x + x^2 + x^3 + \dots + x^{n-1}$ は等比数列の和なので、

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

したがって、

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

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

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

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

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

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

3. 最終的な答え

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

または

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

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

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

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

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

ここで、$x \neq 1$ とする。

最終的な答え：

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

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

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

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

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

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

$S=\frac{1+2x +(3x -5+2) -(3+6+3+...3)}{3*x^2}$

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

$S=\frac{(3)}{x}=x^{-1}$

修正:

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

ここで等比数列の和は $x + x^2 + ... + x^{n-1} = \frac{x(1-x^{n-1})}{1-x} = \frac{x-x^n}{1-x}$.

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

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

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

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

最終的な答え:

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