What is the sum of the squares of the first n numbers?

Question

I think i'm having a brain fart. Can someone give me the formula for this series?

1^2 + 2^2 + 3^2 + ... + (n-1)^2 + n^2

Answer 1

1^2+2^2+3^2+.....+n^2 = n*(n+1)*(2n+1)/6

PS: try google, before posting here

Answer 2

12

Answer 3

There is a formula for the sum of the first n integers to any integral power, but its kind of involved. Here is a link with some formulas for low powers and the general formula http://mathworld.wolfram.com/FaulhabersFormula.html

Answer 4

The sum of 1² through n² is given by
S = (n/6)*(n+1)*(2n+1)

There's an elegent canonical form for sums of the first n integers raised to any integer power that involves the Bernoulli numbers, but I don't remember it off the top of my head. You should be able to find it on the web.


Doug

EDIT: Found it at http://www.mathpages.com/home/kmath279.htm

Answer 5

Here's how to get any such formula: for the nth powers of the first m integers, assume a polynomial in m of degree n, with n unknown coefficients. Use n specific examples to get n equations in these coefficients, and solve them. This always works.