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Initially, I have the following problem: find $$\sum_{k=0}^{n+1}(āˆ’1)^{nāˆ’k}4^k{n+k+1 \choose 2k}.$$ I thought, if I found the function $g_n(x) = \sum_{k=0}^{n}{n+k \choose 2k}x^k$, the answer would be $(-1)^ng_{n+1}(-4)$. I tried to factorize ${n + k \choose 2k}$ to ${n+k \choose k}{n \choose k}/{2k \choose k}$. It is known that $$(1-x)^{-n-1} = \sum_{k=0}^{\infty}{n + k \choose k}x^k,$$ $$(1-4x)^{-1/2} = \sum_{k=0}^{\infty}{2k \choose k}x^k,$$ $$(1+x)^n = \sum_{k=0}^n{n \choose k}x^k.$$ I have tried to combine these formulae, but nothing interesting has been found yet.

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    Have you tried out some small values of $n$ to see if there's a pattern? – 2011-12-27
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    (It looked to me like there was an obvious pattern starting with $n=0,1,2$.) – 2011-12-27

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