How can I evaluate \sum_{n=0}^\infty (n+1)x^n

How can I evaluate

∑n=1∞2n3n+1

I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before so I feel that there must be a simpler method.
In general, how can I evaluate

∑n=0∞(n+1)xn?

homework sequences-and-series convergence power-series

 

 

ANSWER:-

No need to use Taylor Series, this can be derived in a similar way to the formula for geometric series. Lets find a general formula for the following sum:

Sm=∑n=1mnrn.

Notice that

Sm−rSm=−mrm+1+∑n=1mrn

=−mrm+1+r−rm+11−r=mrm+2−(m+1)rm+1+r1−r.

Hence

Sm=mrm+2−(m+1)rm+1+r(1−r)2.

This equality holds for any r, but in your case we have r=13 and a factor of 23 in front of the sum. That is

∑n=1∞2n3n+1=23limm→∞m(13)m+2−(m+1)(13)m+1+(13)(1−(13))2

=23(13)(23)2=12.

Added note:
We can define

Skm(r)=∑n=1mnkrn.

Then the sum above considered is S1m(r), and the geometric series is S0m(r). We can evaluate S2m(r) by using a similar trick, and considering S2m(r)−rS2m(r). This will then equal a combination of S1m(r) and S0m(r) which already have formulas for. This means that given a k, we could work out a formula for Skm(r), but can we find Smk(r) in general for any k? It turns out we can, and the formula is similar to the formula for ∑mn=1nk, and involves the Bernoulli Numbers. In particular, the denominator is (1−r)k+1.