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:
$$\sum_{n=2}^{\infty}(n^2-1)x^n = \sum_{n=1}^{\infty}(n^2-1)x^n =\sum_{n=1}^{\infty}n^2x^n - \sum_{n=1}^{\infty}x^n = \frac{x(1+x)}{(1-x)^3}-\frac{x}{1-x}$$?
 A: 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*}
A: If you check your $\frac{x(1+x)}{(1-x)^3}-\frac{x}{1-x}$ you will find it is $3x^2+8x^3+15x^4 + \cdots$ so you need to remove a factor of $x$ to get  $\frac{(1+x)}{(1-x)^3}-\frac{1}{1-x}$ though I would not write it that way
If you know the coefficient of $x^n$ is $n(n+2)$ then one approach could be to manipulate $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$ and say

*

*$\frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+\cdots$

*$\frac d{dx} \frac{1}{1-x}=1+2x+3x^2+4x^3+5x^4+\cdots$

*$x \frac d{dx} \frac{1}{1-x}=x+2x^2+3x^3+4x^4+5x^5+\cdots$

*$\frac d{dx}\left(x \frac d{dx} \frac{1}{1-x}\right)=1+ 4x+9x^2+16x^3+25x^4+\cdots$

*$\frac d{dx}\left(x \frac d{dx} \frac{1}{1-x}\right) -\frac{1}{1-x} = 3x+8x^2+15x^3+24x^4+\cdots$
which seems to be what you want.
You can then find $$\frac d{dx}\left(x \frac d{dx} \frac{1}{1-x}\right) -\frac{1}{1-x} = \frac{2 x}{{{\left( 1-x\right) }^{3}}}+\frac{1}{{{\left( 1-x\right) }^{2}}}-\frac{1}{1-x}= \frac{x(3-x)}{(1-x)^3}$$
