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I got the following result from WolframAlpha. For positive integer $k$, $$\sum_{i=1}^n i^{2k} C_{2n}^{n-i} = 2^{2n-k-1}\left((2k-1)!!n^k+o(n^k)\right),$$ $$\sum_{i=1}^n i^{2k-1} C_{2n}^{n-i} = \frac12 C_{2n}^{n-1}\left((k-1)!n^k+o(n^k)\right),$$ as $n \to\infty$.

I try for $k=1,\dots,10$, and it is correct. I'm wondering if there is proof of this.

Update: See the proof for $k=1$ in prove a binomial identity for both formulas.

lsstat
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  • Presumably $C_{2n}^{n-i}$ the binomial coefficient $\binom{2n}{n-i}?$ – Thomas Andrews Jan 02 '23 at 14:46
  • Yes, it is the binomial coefficient. – lsstat Jan 02 '23 at 14:49
  • Some possible ideas: First, use the identity by @ Rezha Adrian Tanuharja, see the link: https://math.stackexchange.com/questions/4614246/identity-for-difference-of-sum-of-products-of-binomial-coefficients. Second, the generating function of $i^m\binom{2n}{n-i}$ is $i^m\binom{0}{-i} _3 F_2(\frac12,1,1;1-i,1+i;4z)$ by WolframAlpha, and the partial sum of hypergeometric function may help to prove, see page 223 section 5.7 in the book: csie.ntu.edu.tw/~r97002/temp/Concrete%20Mathematics%202e.pdf. – lsstat Jan 09 '23 at 03:52

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Problem Solved. These two equations have been well-studied in this paper: https://arxiv.org/pdf/math/0606080.pdf. Both polynomials $P_k(n)$ and $Q_k(n)$ have explicit form.

lsstat
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