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The equations $$ \sum_{k=1}^m\begin{pmatrix} n \\ k \end{pmatrix}^2 \quad \text{and} \quad \sum_{k=1}^m\left[2^k\begin{pmatrix} n \\ k \end{pmatrix}\right]^2$$ popped up in some of my calculations, and I was wondering, if there is an elegant solution to it. $\begin{pmatrix} n \\ k \end{pmatrix}$ is the binomial coefficient. The only thing, I found so far is that $$ \sum_{k=1}^n\begin{pmatrix} n \\ k \end{pmatrix}^2 = \begin{pmatrix} 2n \\ n \end{pmatrix} -1, $$ but in my case, $m<n$. Thank you for any leads!

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  • $\begingroup$ In quoting what you found, $k$ should start at $0$. $\endgroup$ – J.G. Apr 15 at 11:21
  • $\begingroup$ Fixed that, thank you! $\endgroup$ – Gemeis Apr 15 at 11:44

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