(2 + 4 + 6 + ... + 20) = ?

1.	(2 + 4 + 6 + ... + 20) = ?
Answer» (22 + 42 + 62 + ... + 202) = (1 x 2)2 + (2 x 2)2 + (2 x 3)2 + ... + (2 x 10)2 = (22 x 12) + (22 x 22) + (22 x 32) + ... + (22 x 102) = 22 x [12 + 22 + 32 + ... + 102] Ref: (12 + 22 + 32 + ... + n2) = 1 n(n + 1)(2n + 1) 6 = 4 x 1 x 10 x 11 x 21 6 = (4 x 5 x 77) = 1540.

Discussion