# Evaluating a summation

I am trying to solve a homework question in which a part involves the evaluation of a summation.

The summation is: $$\sum_{i=0}^n2^{2i+1}$$

and this is my attempt which i am stuck at. Any lead into this would be help full.

$$\sum_{i=0}^n2^{2i+1} = \sum_{i=0}^n (2^{2i})(2) = 2\sum_{i=0}^n2^{2i}$$

when i expand this out like so $$2\sum_{i=0}^n2^{2i} = (2)(1+2^2+2^4+2^6 + ... + 2^{2n})$$

I don't see or know any existing summation that can help me solve this or any way to use pertubation to solve this. Again any lead into helping me out will go a long way

Hint 1: $2^{2i} = 4^i$

Hint 2: $(1+x+x^2+...+x^n)(1-x) = 1-x+x-x^2+x^2-x^3+...+x^n-x^{n+1}$

• ...how stupid of me. such an insult to my intelligence. Oh well thanks a lot anyways! – Krimson Mar 31 '13 at 4:56

Hint: Use the steps you have and the identity $$(1-x)(1+x+x^2+\cdots+x^n)=1-x^{n-1}\Longrightarrow \sum_{i=0}^n x^i=\frac{1-x^{n-1}}{1-x}.$$

This is a geometric series of the form

$$\sum_{k=0}^n r^n = \frac{r^{n+1}-1}{r-1}$$

In your case, $r=2^2=4$.

You could have a direct calculation by using the index of the summation. Let $$S=\sum_{i=0}^n2^{2i+1}=2\sum_{i=0}^n2^{2i}.$$ We have $$2S=2\sum_{i=0}^n2^{2i+1}=\sum_{i=0}^n2^{2i+2}=\sum_{i=1}^{n+1}2^{2i}=\sum_{i=0}^n2^{2i}-1+{2}^{2n+2}=\frac{S}{2}-1+2^{2n+2}.$$ Solve the equation with variable $S$ we obtain $$S=\frac{2}{3}(2^{2n+2}-1).$$