2^2 + 4^2 + 6^2... 20^2 =
(1 * 2)^2 + (2 * 2)^2 + (3 * 2)^2 +..... + (10 * 2)^2 =
2^2 [ 1^2 + 2^2 + 3^2 + .....+ 10^2] =
4 [ n * (n + 1) *(2n + 1) ] / 6 =
(2/3) [ 10 * 11 * 21] =
1540
Note that: \(\boxed{\color{BurntOrange}{1^2 + 2^2 + 3^2 + 4^2 + ... + n^2=\dfrac{n(n+1)(2n+1)}{6}}}\)
\(\color{BurntOrange}{2^2 + 4^2 + 6^2 + ... + 20^2\\ =(1\times 2)^2 + (2\times 2)^2 + (3\times 2)^2 + ... + (10\times 2)^2\\ =1^2\times 2^2 + 2^2\times 2^2 + 3^2\times 2^2 + ... + 10^2 \times 2^2\\ = 2^2(1^2 + 2^2 + 3^2 + ... + 10^2)\\ =2^2\left(\dfrac{(10)(11)(21)}{6}\right)\\ =\left(\dfrac{1}{3}\right)(2)(10)(11)(21)\\ =(7)(2)(10)(11)\\ =(14)(10)(11)\\ =(154)(10)\\ =1540}\)