Closed form of geometric series $\sum_{i=1}^n p^{i+1}$

Question

I know that $\sum_{i=1}^n p^i = \frac{p-p^{n+1}}{1-p}$, but I am not sure how the i+1 factors into the closed form for $\sum_{i=1}^n p^{i+1}$, what is the closed form for the second sum?

Answer 1

You basically answered your question yourself.

You know, $\sum_{i=1}^n p^i = \frac{p \left(p^n-1\right)}{p-1}$

Now

$$\sum_{i=1}^n p^{i+1} = p \sum_{i=1}^n p^i$$

So you get $$\sum_{i=1}^n p^{i+1} = \frac{p^2 \left(p^n-1\right)}{p-1}$$