Using power series representation, calculate ∑n=1∞n2n3n

Recall that, in general,
${\displaystyle 1+x+x^2+\dots =\frac{1}{1-x},\left|x\right|<1}$
Moreover, power series can be differentiated term by term. So, differentiating both sides of the equation above we get
${\displaystyle 1+2x+3x^2+\dots =\frac{1}{{(1-x)}^2},\left|x\right|<1}$
Now, multiplying both sides by x leads to
${\displaystyle x+2x^2+3x^3+\dots =\sum _{n=1}^{\mathrm{\infty }}n{x}^n=\frac{x}{{(1-x)}^2},\left|x\right|<1}$
However, in this case ${\displaystyle x=\frac{2}{3}<1}$, so simply substitute ${\displaystyle x=\frac{2}{3}}$ in formula above.

Basically you start with
${\displaystyle \sum _{x=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}}$
And then you do all the mathematical operations such as ${\displaystyle \frac{d}{dx}}$ on both sides until you get the form you want. For example, the first derivative will give you
${\displaystyle \sum _{x=1}^{\mathrm{\infty }}n{x}^{n-1}=\frac{1}{{(1-x)}^2}}$
A popular second step you can do from there is multiply both sides by x, which gives you
${\displaystyle \sum _{x=1}^{\mathrm{\infty }}n{x}^n=\frac{x}{{(1-x)}^2}}$

First observe that your series is the special case of
$\sum _{n=1}^{\mathrm{\infty }}n{z}^n$
with $z=\frac{2}{3}$, which has radius of convergence R=1
By using the Cauchy product on $\sum _{n=0}^{\mathrm{\infty }}{z}^n=\frac{1}{1-z}$ we get
$(\frac{1}{1-z}{)}^2=(\sum _{n=0}^{\mathrm{\infty }}{z}^n{)}^2=\sum _{n=0}^{\mathrm{\infty }}(\sum _{k=0}^n)z^n=\sum _{n=0}^{\mathrm{\infty }}(n+1)z^n=\sum _{n=1}^{\mathrm{\infty }}n{z}^{n-1}$
and after multiplying by z
$\sum _{n=1}^{\mathrm{\infty }}n{z}^n=\frac{z}{(1-z{)}^2}$
For $z=\frac{2}{3}$ we get $\frac{\frac{2}{3}}{(1-\frac{2}{3}{)}^2}=2\cdot 3=6$