Given that $1^{2}+ 2^{2} + 3^{2} + ... + n^{2}$ = $\frac{n}{6}(n+1)(2n+1)$ ,then $10^{2}+ 11^{2} + 12^{2} + ... + 20^{2}$ is equal to :

Given that $1^{2}+ 2^{2} + 3^{2} + ... + n^{2}$ = $\frac{n}{6}(n+1)(2n+1)$ ,then $10^{2}+ 11^{2} + 12^{2} + ... + 20^{2}$ is equal to :