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

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?

• $\sum_{i=1}^n p^{i+1}=p \sum_{i=1}^n p^{i}$, But your formula is wrong. $\sum_{i=1}^n p^{i}=\frac{p-p^{n+1}}{1-p}$ Nov 3 '20 at 19:46
• Thank you I will update it, so the +1 doesn't change the closed form? Nov 3 '20 at 19:49
• It changes... by muliplying it by $p$ Nov 3 '20 at 19:53
• so it is $\frac{p^2 - p^{n+2}}{1-p}$ ? Nov 3 '20 at 19:55
• thank you I will create an answer for the question. Nov 3 '20 at 19:58

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