Find the sum of the series $3x+8x^2+15x^3 + ....$ I'm trying to express the following series $3x+8x^2+15x^3 + ....$ as a sum and hope to find its sum for $|x| < 1$
Here is what I have so far:
To me the series looks to be the derivative of the following form:
$1 + \frac{3}{2}x^2+\frac{8}{3}x^3+\frac{15}{4}x^4 ...$
Given that the denominator progresses as $\frac{1}{n}$ I had thought of taking it out like so:
$\frac{1}{n}(1+3x^2+8x^3+15x^4 ....)$
Now I'm unsure of the sequence that the constants follow but lets denote this as $a$ and take it out of the sequence of $x$ values, then I get:
$\frac{1}{n}(1+a(x^2+x^3+x^4+x^5 ...))$
We know that the sum of the series of $x$ follows the following geometric series: $\frac{x^2}{(1-x)}$, plugging this into the equation:
$\frac{1}{n}(1+\frac{ax^2}{1-x})$
Then taking the first derivative I get:
$\frac{ax(2+x)}{n(1-x)^2}$
I'm unsure as to whether the $n$ can still be introduced or whether it's removed - How is this answer optimally derived?
Following the hint below:
$$\sum_{n=2}^{\infty}(n^2-1)x^n = \sum_{n=1}^{\infty}(n^2-1)x^n =\sum_{n=1}^{\infty}n^2x^n - \sum_{n=1}^{\infty}x^n = \frac{x(1+x)}{(1-x)^3}-\frac{x}{1-x}$$?

Hint $1$: First notice that the coefficients are $n^2-1$.
Hint $2$:
\begin{eqnarray*}
 \sum_{n=1}^{\infty}  n^2 x^n  =\frac{x(1+x)}{(1-x)^3}.
\end{eqnarray*}
