How can I calculate the sum of $\sum\limits_{n = 1}^{\infty} n(n+3)x^n$?

Question

How can I calculate the sum of $\sum\limits_{n = 1}^{\infty} n(n+3)x^n$ analytically?

Answer 1

First you split your series into two series using the fact that $$n(n+3) = n(n+1)+2n$$

Then note that derivative of $x^n$ is $nx^{n-1}$ so $$nx^n = x(\frac {d}{dx}(x^n))$$

Then note that the second derivative of $x^{n+1}$ is $n(n+1)x^{n-1}$ and go from there to write your series as some alterations of geometric series.

Answer 2

Hint:

$\displaystyle\sum\limits_{n = 1}^{\infty} n(n+3)x^n=x^2\sum_{n = 1}^{\infty}n(n-1)x^{n-2}+4x\sum_{n = 1}^{\infty}nx^{n-1}=x^2\Bigl(\sum_{n = 1}^{\infty}x^n\Bigr)''+4x\Bigl(\sum_{n = 1}^{\infty}x^n\Bigr)'$.

Answer 3

if $-1< x<1$ $$ \\\sum_{n=1}^{+\infty}n(n+3)x^n=x\cdot\frac{d}{dx}\sum_{n=1}^{+\infty}(n+3)x^n=x\frac{d}{dx}S(x) \\S(x)=\sum_{n=1}^{+\infty}(n+3)x^n=x\cdot\sum_{n=1}^{+\infty}n\cdot x^{n-1}+3\sum_{n=1}^{+\infty}x^n= \\=x\frac{d}{dx}\sum_{n=1}^{+\infty}x^n+\frac{3x}{1-x}=x\frac{d}{dx}\frac{x}{1-x}+\frac{3x}{1-x} => \\S(x)=x(\frac{3}{1-x}+\frac{1}{(1-x)^2})=> \\\sum_{n=1}^{+\infty}n(n+3)x^n=x\cdot\frac{d}{dx}S(x) $$