$\sum_{n=1}^\infty n*(n+1)*x^n$
Hello everyone, I need help in solving the question above.
I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don't know how to continue. Thanks.
On
Just to make the problem more general.
Consider $$\sum_{n=1}^\infty (an^2+bn+c)\,x^n$$ rewrite $n^2=n(n-1)+n$ which makes $$an^2+bn+c=a n(n-1)+a n+bn+c=an(n-1)+(a+b)n+c$$ $$\sum_{n=1}^\infty (an^2+bn+c)\,x^n=a \sum_{n=1}^\infty n(n-1)x^n+(a+b)\sum_{n=1}^\infty nx^n+c\sum_{n=1}^\infty x^n$$ $$\sum_{n=1}^\infty (an^2+bn+c)\,x^n=a x^2 \sum_{n=1}^\infty n(n-1)x^{n-2}+(a+b)x\sum_{n=1}^\infty nx^{n-1}+c\sum_{n=1}^\infty x^n$$ $$\sum_{n=1}^\infty (an^2+bn+c)\,x^n=a x^2\left(\sum_{n=1}^\infty x^n \right)''+(a+b)x\left(\sum_{n=1}^\infty x^n \right)'+c\left(\sum_{n=1}^\infty x^n \right)$$
If we had an $n^3$ term, the same idea $$n^3=n(n-1)(n-2)+3n(n-1)+n$$ and so on for $n^k$.
Using $$\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$$ then $$\sum_{n} n \, x^{n+1} = x^2 \, \frac{d}{dx} \, \frac{1}{1-x}$$ and $$\sum_{n} n \, (n+1) \, x^n = \frac{d}{dx} \left( x^{2} \, \frac{d}{dx} \frac{1}{1-x} \right).$$
After some work it is found that: \begin{align} \sum_{n=0}^{\infty} x^{n} &= \frac{1}{1-x} \\ \sum_{n=0}^{\infty} n \, x^{n} &= \frac{x}{1-x} \\ \sum_{n=0}^{\infty} n(n+1) \, x^{n} &= \frac{2 x \, (x^2 - x + 1)}{(1-x)^{3}} = \frac{2 x (1+x^3)}{(1-x)^2 \, (1-x^2)}. \end{align}