KöMaL 2024 B5411

Prove the following statement for every $n\ge 2$ positive integer: $\sum_{k=1}^{n^2-1} \lfloor \sqrt{k} \rfloor]^2=\frac{(n-1)n(3n^2-n-1)}{6}$. $\lfloor x \rfloor$ denotes the floor of the real number $x$.