how to find power series in closed form

Question

Find a "closed form" of summation $n=2$ to infinity $n(n-1)x^n$.

I don't have much clue to solve this. Can anyone please explain how to approach this?

Answer 1

A start: We have for suitable $x$ $$\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots.$$ Differentiate twice.

Answer 2

Just rephrasing what André Nicolas wrote, consider the term $$u_n=n(n-1)x^n=x^2\Big(n(n-1)x^{n-2}\Big)$$ I am sure that you recognize that what is inside the parentheses is the second derivative of $x^n$. So, $$\sum_{n=2}^{\infty} u_n=x^2\frac{d^2}{dx^2}\Big(\sum_ {n=2}^{\infty} x^n\Big)$$ and in the rhs, you have a geometric progression; so, provided a suitable $x$, $$\sum_ {n=2}^{\infty} x^n=\frac{x^2}{1-x}$$ Taking derivatives and simplifying leads to $$\frac{d^2}{dx^2}\Big(\sum_ {n=2}^{\infty} x^n\Big)=\frac{2}{(1-x)^3}$$ and, finally,$$\sum_{n=2}^{\infty} u_n=\frac{2 x^2}{(1-x)^3}$$