Question 6
Prove that
$$\sum_{n=0}^{2013}\frac{4026!}{(n!(2013-n)!)^2}$$
is the square of an integer.
$$\sum_{n=0}^{2013}\frac{4026!}{(n!(2013-n)!)^2}$$
is the square of an integer.
MathNet 2013 Question 6
Math Olympiad Vault · MathNet 2013
Math Olympiad Vault · MathNet 2013