The sum . What is the value of ?

Awesomeee's question is 21^2+22^2+...+40^2

not 1^2+2^2+3^2+4^2+....39^2+40^2

21^2+22^2+...+40^2=[(1^2+2^2+3^2+...39^2+40^2)-(1^2+2^2+...19^2+20^2)]/6

                                = $${\frac{\left[{\mathtt{40}}{\mathtt{\,\times\,}}\left({\mathtt{40}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}\left({\mathtt{81}}\right){\mathtt{\,-\,}}{\mathtt{20}}{\mathtt{\,\times\,}}\left({\mathtt{20}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}{\mathtt{41}}\right]}{{\mathtt{6}}}} = {\mathtt{19\,720}}$$

Thanks, fiora....I totally mis-read that one.....!!!!