Prove closed form for $\sum_k (2k)^2\binom{2n}{k+n} $ and $\sum_ k(2k+1)^2\binom{2n+1}{k+n+1}$

Question

The following identities, are true, but I am having trouble proving them: $$ \left\{\begin{array}{lll} \sum_k (2k)^2\binom{2n}{k+n} &=& (2n)2^{2n} \\ \sum_ k(2k+1)^2\binom{2n+1}{k+n+1} &=& (2n+1)2^{2n+1} \end{array} \right. $$ The sums in these identities range over all integer values of $k$, but for any given $n$ there are just a finite number of non-zero terms. [The convention (following Knuth) is that a binomial coefficient with a negative lower index like $\binom{7}{1}$ or an lower endex exceeding the upper index like $\binom{8}{9}$ is zero.]

These identities can be used to show that the expectation of $x^2$ at epoch $t$ of a 1-dimensional discrete $p=\frac{1}{2}$ random walk along the X axis is $t$.

I have tried induction on $n$, but the algebra has gotten a bit hairy. Is there a cleaner way? And if not, can somebody demonstrate the induction step (the basis is easy for $n=1$)?

Answer 1

These can both be proven using the binomial identity $$ \binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c}\tag{1} $$ For the first $$ \begin{align} \sum_{k=-n}^n(2k)^2\binom{2n}{k+n} &=4\sum_{k=0}^{2n}(k-n)^2\binom{2n}{k}\\ &=4\sum_{k=0}^{2n}\left[2\binom{k}{2}-(2n-1)\binom{k}{1}+n^2\right]\binom{2n}{k}\\ &=4\sum_{k=0}^{2n}\left[2\binom{2n-2}{k-2}\binom{2n}{2}-(2n-1)\binom{2n-1}{k-1}\binom{2n}{1}+n^2\binom{2n}{k}\right]\\[6pt] &=4\left[2n(2n-1)2^{2n-2}-2n(2n-1)2^{2n-1}+n^22^{2n}\right]\\[12pt] &=4\left[-n(2n-1)2^{2n-1}+2n^22^{2n-1}\right]\\[12pt] &=2n2^{2n}\tag{2} \end{align} $$

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For the second $$ \begin{align} \sum_{k=-n}^n(2k+1)^2\binom{2n+1}{k+n+1} &=\sum_{k=0}^{2n}(2k-2n+1)^2\binom{2n+1}{k+1}\\ &=\sum_{k=0}^{2n}\left[8\binom{k+1}{2}-8n\binom{k+1}{1}+(2n+1)^2\right]\binom{2n+1}{k+1}\\ &=\sum_{k=0}^{2n}\left[8\binom{2n+1}{2}\binom{2n-1}{k-1}-8n\binom{2n+1}{1}\binom{2n}{k}+(2n+1)^2\binom{2n+1}{k+1}\right]\\[6pt] &=\left[8n(2n+1)2^{2n-1}-8n(2n+1)2^{2n}+(2n+1)^22^{2n+1}\right]\\[12pt] &=\left[-8n(2n+1)2^{2n-1}+4(2n+1)^22^{2n-1}\right]\\[12pt] &=(2n+1)2^{2n+1}\tag{3} \end{align} $$

Answer 2

We may consider that:

$$ \sum_{k=-n}^{n}\binom{2n}{n+k} z^k = \frac{(1+z)^{2n}}{z^n} \tag{1}$$ then replace $z$ with $e^{i\theta}$ and take the real part: $$ -\binom{2n}{n}+2\sum_{k=0}^{n}\binom{2n}{n+k} \cos(k\theta) = \left(2\cos\frac{\theta}{2}\right)^n\tag{2}$$ differentiate twice with respect to $\theta$ then consider $\theta\to 0$.

The odd case is similar.

Answer 3

Just shift $k$ to $k-n$, then \begin{align} \sum_{k=-n}^n (2k)^2 \binom{2n}{n+k}&=\sum_{k=0}^{2n}(2(k-n))^2\binom{2n}{k}\\ &=4\left(\sum_{k=0}^{2n}n^2\binom{2n}{k} -\sum_{k=0}^{2n}2nk\binom{2n}{k}+ \sum_{k=0}^{2n}k^2\binom{2n}{k}\right)\\ &=4(n^2 2^{2n}-2n\cdot 2n 2^{2n-1}+2n\cdot 2^{2n-1}+2n(2n-1)2^{2n-2})\\ &=(4n^2-8n^2+4n+4n^2-2n)2^{2n}\\ &=2n\cdot 2^{2n}. \end{align} Using similar method, you can also verify second identity: \begin{align} \sum_{k=-n-1}^n (2k+1)^2 \binom{2n+1}{k+n+1}&=\sum_{k=0}^{2n+1}(2k-2n-1)^2\binom{2n+1}{k}\\ &=(2n+1)2^{2n+1}. \end{align}