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I need help finding a closed form of this finite sum. I'm not sure how to deal with sums that include division in it.

$$\sum_{i=1}^n \frac{2^i}{2^n}$$

Here's one of the attempts I made and it turned out to be wrong:

$$\frac{1}{2^n}\sum_{i=1}^n {2^i} = \frac{1}{2^n} (2^{n +1} - 1) = \frac{2^{n + 1} - 1} {2^n}$$

And then from there, simplifying it ended up with just a constant.

I also tried it in which I moved $2^{-n}$ to the outside of the sigma notation and went from there:

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

I plugged the equation in to Wolfram Alpha to check my answers. It gave me $2 - 2^{1-n}$, which is close to what I got in that second method. I need help finding the error in my math. I keep looking over it and I guess I'm just not seeing something.

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  • $\begingroup$ Both of your attempts are the same, the only difference is how you call $\frac{1}{2^n}=2^{-n}$. Note that $\sum\limits_{i=a}^b x^i = \frac{x^a-x^{b+1}}{1-x}$ $\endgroup$ – JMoravitz Jan 13 '17 at 21:31
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Your first attempt was a good idea but you made some mistakes in your computations.

Since the denominator does not depend on $i$ you can take it out of the sum and you get

$$\sum_{i=1}^n \frac{2^i}{2^n}=\frac{1}{2^n}\sum_{i=1}^n {2^i} = \frac{1}{2^n} 2 (2^{n} - 1) = \frac{2^{n} - 1} {2^{n-1}}=2-2^{1-n}$$

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  • $\begingroup$ I can see where I made my mistake now, thank you for showing me step-by-step! $\endgroup$ – Jasmine Jan 13 '17 at 21:42
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$$\sum _{ i=1 }^{ n }{ 2^{ i } } =\frac { 2\left( 1-{ 2 }^{ n } \right) }{ 1-2 } ={ 2 }^{ n+1 }-2$$

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