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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

Thanks in advance!

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

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

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  • $\begingroup$ ...how stupid of me. such an insult to my intelligence. Oh well thanks a lot anyways! $\endgroup$ – Krimson Mar 31 '13 at 4:56
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Hint: Use the steps you have and the identity $$(1-x)(1+x+x^2+\cdots+x^n)=1-x^{n-1}\Longrightarrow \sum_{i=0}^n x^i=\frac{1-x^{n-1}}{1-x}.$$

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This is a geometric series of the form

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

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

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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 $$ S=\frac{2}{3}(2^{2n+2}-1). $$

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