We want to find a closed form expression for $$\sum_{k=1}^{n}4^k\binom{2n}{2k}$$ I've been trying to come up with some combinatoric approach to this problem. The number of ways to choose some $2k$ elements from a set of $2n$ elements is $$\binom{2n}{2k}$$ but this is about as far as I can get. Any helpful identities or theorems/guidance would be greatly appreciated!
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2Have you evaluated for small values of $n$? If yes, can you figure out what the pattern is? Did you at least try Wolfram? – Calvin Lin Oct 14 '20 at 00:26
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Does this answer your question? Sums of binomial coefficients – Anindya Prithvi Oct 14 '20 at 06:33
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Note that $$\sum_{k\ge0} a_{2k} = \sum_{k\ge0} \frac{1+(-1)^k}{2}a_k.$$ Taking $$a_k=2^k \binom{2n}{k}$$ yields \begin{align} \sum_{k\ge0} 2^{2k}\binom{2n}{2k} &= \sum_{k\ge0} \frac{1+(-1)^k}{2} 2^k \binom{2n}{k} \\ &= \frac{1}{2}\sum_{k\ge0} 2^k \binom{2n}{k} + \frac{1}{2}\sum_{k\ge0} (-2)^k \binom{2n}{k} \\ &= \frac{1}{2}(1+2)^{2n} + \frac{1}{2}(1-2)^{2n} \\ &= \frac{9^n+1}{2}. \end{align} So $$\sum_{k=1}^n 2^{2k}\binom{2n}{2k} = \frac{9^n+1}{2} - 2^0\binom{2n}{0} = \frac{9^n-1}{2}.$$
RobPratt
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Hint : $2^{2k}=4^k$ \begin{eqnarray*} (2+1)^{2n} &=& \sum_{k=0}^{2n} \binom{2n}{k} 2^{k} \\ (2-1)^{2n} &=& \sum_{k=0}^{2n} (-1)^k \binom{2n}{k} 2^k.\\ \end{eqnarray*}
Alexander Burstein
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Donald Splutterwit
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1does binomial theorem not say that $(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$? If so, where does the $\binom{2n}{2k}$ come from? – Anonymous Oct 14 '20 at 00:38
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3Check your summations: $$\sum_{k=0}^{2\cdot1}\binom{2\cdot1}{2k}2^k=\binom202^0+\binom222^1+\binom242^2=3\ne9=(2+1)^{2\cdot1}$$ Presumably you mean to add $\sum_{k=0}^{2n}\binom{2n}k2^k$ and $\sum_{k=0}^{2n}(-1)^k\binom{2n}k2^k$. – Brian M. Scott Oct 14 '20 at 01:10
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In a more general manner $$\sum_{k=1}^{n}\binom{2n}{2k}x^k=-1+\frac{1}{2} \left(\left(\sqrt{x}-1\right)^{2 n}+\left(\sqrt{x}+1\right)^{2 n}\right)$$
Claude Leibovici
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