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Find the sum of: $1×2+ 2×3x+ 3×4x^2...$ I tried the problem and I am getting answer as $\frac{(2-x)}{(1-x)^2}$ which I think is wrong Can Someone please tell the correct answer so that I can find my mistake

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  • $\begingroup$ Why don’t you show your calculations so that we can point out the error? $\endgroup$ – Martin R May 13 at 6:59
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Use the following. $$1\cdot2+2\cdot3x+3\cdot4x^2+...=\left(2x+3x^2+4x^3+...\right)'=(x^2+x^3+x^4+...)''$$ For $-1<x<1$ I got $\frac{2}{(1-x)^3}.$

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$S = 1×2+ 2×3x+ 3×4x^2+4×5x^3...$

$Sx = ~~~~~~~~~~~1×2x+ 2×3x^2+ 3×4x^3...$(multiplying $S$ with $x$)

Subtracting,

$(S-Sx) = 2+4x+6x^2+8x^3+\dots\Rightarrow 2(1+2x+3x^2+4x^3+\dots) = 2L$(suppose L is that sum.)

$L = 1+ 2x+ 3x^2+4x^3+\dots$

$Lx = ~~~~~~x+2x^2+3x^3+\dots$

Subtracting,

$L(1-x) = \frac{1}{1-x} \Rightarrow L= \frac{1}{(1-x)^2}$

Putting the value of $L$ in the above equation, $S(1-x)=2L=2\times\frac{1}{(1-x)^2}$, therefore, $\boxed{S= \frac{2}{(1-x)^3}}$

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The sum is: $$\sum_{i=1}^{\infty} i(i+1)x^{i-1}$$

The standard method is: $$\begin{align}f(x)&=\sum_{i=0}^{\infty}x^{i+1}=\frac{x}{1-x},|x|<1 \Rightarrow \\ f'(x)&=\sum_{i=0}^{\infty} (i+1)x^i=\frac{1}{(1-x)^2} \Rightarrow \\ f''(x)&=\sum_{i=1}^\infty i(i+1)x^{i-1}=\frac{2}{(1-x)^3}.\end{align}$$

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