# Reason what 2^k in a sigma will do

I am trying to solve the following calculation without a calculator:

$$\sum_{k=0}^82^k{8\choose k}$$

The first part:

$$\sum_{k=0}^8{8\choose k}$$

is equal to $2^8$. I already know that the answer will be $3^8$. How did the $2^k$ transform the answer from $2^8$ to $3^8$?

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Use that $$(a+b)^n = \sum_{k=0}^n a^k b^{n-k} \binom{n}{k} \text{.}$$ For $a=2$, $b=1$ you get the sum you want to compute, and the result is therefore $(2+1)^8$

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$$(x+1)^8= \binom{8}{0}+\binom{8}{1}x+\binom{8}{2}x^2+\cdots+\binom{8}{8}x^8$$

Let $x=2$ and we are done.

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Sorry ! Typo's. – Claude Leibovici Apr 6 '14 at 14:19

Hint: Take a look at the proof of why $$\sum_{k=0}^8{8\choose k}$$ is equal to $2^8$. I presume that it was done using the binomial theorem. Using the same theorem, try to modify it so you will get the $2^k$ values in the expansion.

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@DavidRicherby I was sloppy in my writing and wanted to write too much too quickly. I edited my answer, thank you. – 5xum Apr 7 '14 at 6:54