Given that 1-x-x2- - xn-1 - 1-xn-1-x- x - 1- Using this- find the sum of the series 1-2x-3x2- - n-1xn-2

1+x+x^2+⋯ +x^n 1=1 x^n1 x Differentiating on both sides with respect to x 1+2x+3x^2+⋯+(n 1)x^n 2=(1 x)( nx^n 1)+1 x^n(1 x)^2=(n 1)x^n nx^n 1+1(1 ...

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AM,GM,HM Inequality

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Question

Given that $1 + x + x^{2} + \dots x^{n - 1} = \frac{1 - x^{n}}{1 - x}, x \neq 1$ , Using this, find the sum of the series $1 + 2 x + 3 x^{2} + \dots (n - 1) x^{n - 2}$

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1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . .

$The value of l i m x \to 1 \frac{x^{n} + x^{n - 1} + x^{n - 2} + . . . . . . . + x^{2} + x - n}{x - 1}$

1 + 2 x + 3 x^{2} + 4 x^{3} + . . . . .

(1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) +

r \geq 1

x \neq 1

t_{r} = 1 + 2 x + 3 x^{2} + \dots + r x^{r - 1}

t_{1} + t_{2} + \dots + t_{n}

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

r \geq 1

x \neq 1, 1 + x + x^{2} + \dots + x^{r - 1} = \frac{1 - x^{r}}{1 - x}

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

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