Derive an explicit formula for a power series $\sum_{n=1}^\infty n^2x^n$ Could anyone help me find an explicit formula for:

$$
\sum_{n=1}^\infty n^2x^n
$$

We're supposed to use: $$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}  \qquad  |x| &lt;1 $$
We know that the power series for:

$$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}  \qquad  |x| &lt;1 $$

In order to find the power series for:

$$
\sum_{n=1}^\infty n^2x^n
$$

We differentiate the first summation above:

$$\left(\sum_{n=1}^\infty nx^n\right)' = \sum_{n=1}^\infty n^2x^{n-1} =  -\frac{x+1}{{(x-1)}^3}  \qquad  |x| &lt;1$$

Finally, we multiply by $x$:

$$x\sum_{n=1}^\infty n^2x^{n-1} = \sum_{n=1}^\infty n^2x^{n} = -\frac{x(x+1)}{{(x-1)}^3}$$
Note that if $$s(x)=\sum f(n)x^n$$ then $$xs'(x)=\sum nf(n)x^n$$

Apply with $f(n)=1$, and then with $f(n)=n$. Just take care with the lower limit and the early terms.