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Find: $\sum_{k=0}^n \binom{n}{k}2^k$

So I know how to find the coefficients of binomial expansions but I’m not sure how to do this type of problem where I just find the summation.

The example in the book is $\sum_{k=0}^n \binom{n}{k} 4^k $ and the answer is $5^n$

Thank you for any and all help.

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  • $\begingroup$ I’m sorry I’m still new to this and using mathjax as well thank you again. $\endgroup$
    – C.Math
    Feb 16, 2018 at 16:51
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    $\begingroup$ Use the binomial theorem with $(1+2)^n$. In the example in the book, didn't they expand $(1+4)^n$? $\endgroup$
    – user530891
    Feb 16, 2018 at 16:52
  • $\begingroup$ By example I meant just that there was another question and the back of the book had the answer $5^n$ $\endgroup$
    – C.Math
    Feb 16, 2018 at 16:55
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    $\begingroup$ Both can be solved that way. $\endgroup$
    – user530891
    Feb 16, 2018 at 16:56

2 Answers 2

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$$\sum_{k=0}^n \binom{n}{k}2^k =\sum_{k=0}^n \binom{n}{k}2^k 1^{n-k}=(2+1)^n$$ By Binomial theorem.

$$(a+b)^n=\sum_{k=0}^n \binom{n}{k}a^kb^{n-k}$$

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  • $\begingroup$ So all the question is looking for is an answer of the expansion $(1+2)^n$ or $3^n$? $\endgroup$
    – C.Math
    Feb 16, 2018 at 16:58
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    $\begingroup$ Given the expansion, find the compact form. Yes, just like $5^n$ is the answer to the other question, $3^n$ is the answer to this question. You are being tested on the ability to recognize the formula. $\endgroup$
    – Siong Thye Goh
    Feb 16, 2018 at 17:00
  • $\begingroup$ Ah I understand now. Thank you for your help! $\endgroup$
    – C.Math
    Feb 16, 2018 at 17:18
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Hint:

rewrite is as $$\sum_{k=0}^n\binom{n}{k}2^k1^{n-k}$$ Any recognition?

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