Calculate $\sum^{\infty}_{n=1}n^2x^{n-1}$

Question

I have some struggles with this exercise. I need to find out $$\sum_{n=1}^{\infty} n^2x^{n-1}$$ I know that the answer is $\frac{1+x}{(x-1)^3} $ when $|x|<1$ $ $. And I need to solve it by using integration and derivatives. But when I do so I get the answer $\frac{1}{1-x}$...

Possible duplicate of How to fit $\sum{n^{2}x^{n}}$ into a generating function? — Sil, Apr 19 '17 at 20:24
Not exact duplicate but the difference is only dividing by $x$ and @Gerry Myerson answer there is what you need i think... — Sil, Apr 19 '17 at 20:25
This one is also similar How to prove that $\sum_{n=1}^\infty{n^2a^{n-1}}=\frac{1+a}{(1-a)^3}$ — Sil, Apr 19 '17 at 20:31
And this one Calculate $\sum_{n=1}^{\infty}n^2q^{n-1}$ ... Quite a popular question :) — Sil, Apr 19 '17 at 20:32
You can probably find several similar questions using Approach0 — Martin Sleziak, Apr 20 '17 at 01:56

score 6 · Accepted Answer · answered Apr 19 '17 at 20:25

6

Hint: $$ \sum_{n=1}^\infty n^2x^{n-1}=x\sum_{n=1}^\infty n(n-1)x^{n-2}+\sum_{n=1}^\infty nx^{n-1} $$

answered Apr 19 '17 at 20:25

operatorerror

29,103

Calculate $\sum^{\infty}_{n=1}n^2x^{n-1}$

1 Answers1