What is the value of summation $\sum_{k=0}^{n}C(2n+1,k)$ ?
Assume $C(n,k)$ implies to choose $K$ items out of $N$ items
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1What is $C(2n+1, k)$? is it the choose function? – S.C.B. Feb 28 '17 at 02:58
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If $C(2n+1, k)=\binom{2n+1}{k}$ then the sum evaluates to $4^n$ – Dando18 Feb 28 '17 at 03:01
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Hint: use $\binom{2n+1}{k}=\binom{2n+1}{2n+1-k}$.
$$2\sum_{k=0}^n \binom{2n+1}{k} = \sum_{k=0}^n \left(\binom{2n+1}{k} + \binom{2n+1}{2n+1-k}\right) = \sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}$$
angryavian
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1@JonGarrick It doesn't matter; you could just as well take everything I wrote and divide it by $2$. I just did it that way because I knew that the sum $\sum_{k=0}^n$ would be "half" of a full sum $\sum_{k=0}^{2n+1}$. – angryavian Feb 28 '17 at 04:29
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