An algebra problem by ashutosh malviya

$\color{#20A900}{1^2+2^2+3^2+\dotsm+n^2=\frac{n(n+1)(2n+1)}{6}}$ $\color{#3D99F6}{1^2+2^2+3^2+\dotsm+20^2=\frac{20(20+1)(2(20)+1)}{6}=2870}$ $\color{#D61F06}{5^2+6^2+7^2+\dotsm+20^2=2870-(1^2+2^2+3^2+4^2)=2870-30=\boxed{2840}}$

$1^2+2^2+\cdots+20^2=\dfrac{20(20+1)(2(20)+1)}{6}=2870$

$5^2+6^2 \cdots+20^2=2870-(1^2+\cdots+4^2)=2870-30=\boxed{2840}$ ​

1+2x2+3x3+...+nxn=n(n+1)(2n+1)/6 so we get 20x21x41/6-4x5x9/6=2870-30=2840

$5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 18^2 + 19^2 + 20^2 = \fbox {2840}$