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How can I find the sum of the following series: $\sum_{n=2}^{\infty} n(n-1)x^{n+1}$?

The answer key of the problem set says the answer is: $\frac{2x^3}{(1-x)^3}$, but I'm not sure how to get that answer.

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    $\begingroup$ Differentiate $\sum x^n$ twice. $\endgroup$ – Gerry Myerson Oct 3 '15 at 2:51
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HINT:

Let $$S_n=\sum_{r=2}^nr(r-1)x^r$$

$$(1-x)S_n=n(n-1)x^n+\sum_{r=2}^{n-1}x^r\{r(r-1)-(r-1)(r-2)\}$$

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

Again, let $$T_m=\sum_{r=2}^m(r-1)x^r$$

$$(1-x)T_m=?$$

Now see this

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We know $\sum _{n=0}^\infty x^n=\frac1{1-x}$ (*). Let's try to play with this to get to our desired expression.

First, in your expression n starts from $2$ instead of $0$. So we change it to:

$\sum_{n=2}^\infty n(n-1)x^{n+1}=\sum_{n=0}^\infty (n+2)(n+1)x^{n+3}$

Now, we observe that $(n+2)(n+1)x^n$ is what we get when we differentiate $x^{n+2}$ twice.

Multiplying (*) by $x^2$:

$\sum _{n=0}^\infty x^{n+2}=\frac{x^2}{1-x}$

Now I leave it to you to differentiate both sides twice, then multiply both sides by $x^3$ (to get $x^{n+3}$ instead of $x^n$) and get your result.

(I did it on paper, I got it right)

Good luck!

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