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sum of consecutive square numbers?
I'm trying to work out a formula for the sum of:
1^2 + 2^2 + 3^2 + 4^2+...+ n^2
so that I could input n and have the formula return the sum. I'm shooting blanks. Anybody have any ideas?
2 Answers
- Roger the MoleLv 71 decade agoFavourite answer
The formula you're looking for is:
n(n+ 1)(2n +1)/6
Unfortunately, the method I used to derive it won't stand inspection. I can only assert it, not prove it.
It looked like a polynomial, so I hypothesized something big:
An^4 + Bn^3 + Cn^2 + Dn + E = sum
Then I took the first five instances of n and the corresponding first five sums, plugged them in, creating 5 equations in 5 unknowns. Solved that system of equations and got:
A= 0, B = 1/3, C = 1/2 , D = 1/6, E = 0.
I put those coefficients back into the original hypothesis, and rearranged to get what I asserted above.
I'm sure there's a much cleverer way, along the lines of Gauss' solution to the sum of consecutive integers from 1 to n, but I'm not seeing it just now.
