Square pyramidal number

Statement

The $n$-th square pyramidal number is the sum of the first $n$ squares: $\mathrm{Pyr}_n=1^2+2^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}{6}$. Compute $\mathrm{Pyr}_6$.

Solution

By the formula: $\mathrm{Pyr}_6=\frac{6\cdot 7\cdot 13}{6}=\boxed{91}.$ Direct check: $1+4+9+16+25+36=91$.