# What is $\sum_{k=1}^m\left[2^k\begin{pmatrix} n \\ k \end{pmatrix}\right]^2$?

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. Thank you for any leads!

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