0

Prove that $\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2=\binom{2n}{n}$ by using a block-walking argument.

I found the identity but I wasn't able to find a block-walking argument. Could I please have some help? Thanks! :D

  • Not sure what you mean by block-walking, but this could help : https://math.stackexchange.com/questions/148583/combinatorial-proof-of-summation-of-sum-limits-k-0n-n-choose-k2-2n – Arnaud D. Apr 11 '19 at 18:37

1 Answers1

2

Every lattice path from $(0,0)$ to $(n,n)$ must pass through exactly one of the circled points. How many paths are there from $(0,0)$ to $(k,n-k)$? How many from $(k,n-k)$ to $(n,n$)?

Mike Earnest
  • 75,930