As the title states, is there any simplification to $\sum\limits_{n=k}^N \binom{n}{k}^2$? I found this which sums the squares of the "rows" of a pascal triangle, but here I'm trying to sum the diagonals. There is also this mathSE question from a few years ago with no answers.
Update: I shall clarify that the answer is not a central binomial coefficient, nor using the hockey-stick identity. This is directly seen as $\binom{3}{3}^2 + \binom{4}{3}^2 + \binom{5}{3}^3 + \binom{6}{3}^2 = 517$ is not a central binomial coefficient.
Best Answer
Let $T(n,k)=\sum_{j=k}^{n}{j \choose k}^2$.
Computing some values and then searching OEIS you can find your function at OEIS A110197. I suppose there isn't a closed form for the general case, because there isn't any there. However, there are links to special cases sequences and corresponding formulas:
$$T(n,0)=n+1$$ $$T(n,1)=n(n+1)(2n+1)/6$$ $$T(n+2,2)=n(n+1)(n+2)(3n^2 + 6n + 1)/60$$
and other cases up to $k=5$.