The sum 1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6.
What is the value of 21^2 + 22^2 + 23^2 + ... + 40^2 + 41^2 + 42^2?
To solve this, we need to find 1^2...42^2 and then subtract 1^2...20^2 from it.
You already know the formula 1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6.
42(42+1)(2*42+1)/6-20(20+1)(2*20+1)/6 = 22715
I hope this helped. :))
=^._.^=
