sequences and series - Calculate $sum_{n=1}^{infty}n^2q^{n-1}$

Please show me how to calculate the sum of this infinite series:

$$\sum_{n=1}^{\infty}n^2 q^{n-1}$$

I should have included the condition $\mid q\mid$<1

And I was able to solve the infinite series of
$$S_n=\sum_{n=1}^{\infty}n q^{n-1}=1+2q+3 q^2+4q^3...$$
The trick is to calculate

$$q S_{n}=\sum_{n=1}^{\infty}n q^{n}=q+2q^2+3q^3+...$$
And find out that
$$S_{n}-q S_{n}=\sum_{n=1}^{\infty}=1+q+q^2+q^3+...=\sum_{n=1}^{\infty}q^n=\frac{1}{1-q}$$
And thus
$$S_{n}=\frac{\frac{1}{1-q}}{1-q}=\frac{1}{(1-q)^2}$$

But I was not able to use same trick on the series I want to solve.

P.S. @user17762 has provided a genius way to handle this kind of series and his approach could simplify the calculation of $\sum_{n=1}^{\infty}n q^{n-1}$ too. Just watch the first two steps he used.

Answer

Your series converges only when $\vert q \vert <1$. Here is one possible way to derive the sum of the infinite series, for $\vert q \vert <1$. We have
$$\sum_{n=1}^{\infty} q^n = \dfrac{q}{1-q} = -1 - \dfrac1{q-1}$$
Differentiate this once to get
$$\sum_{n=1}^{\infty} n q^{n-1} = \dfrac{d}{dq} \left(\dfrac{q}{1-q} \right) = \dfrac1{(q-1)^2}$$
Multiplying by $q$, we get that
$$\sum_{n=1}^{\infty} n q^{n} = \dfrac{q}{(q-1)^2} = \dfrac1{q-1} + \dfrac1{(q-1)^2}$$
Differentiate this again, to get
$$\boxed{\displaystyle \sum_{n=1}^{\infty} n^2 q^{n-1}=-\dfrac1{(q-1)^2} - \dfrac2{(q-1)^3} = \color{blue}{\dfrac{(1+q)}{(1-q)^3}}}$$