Frequently Asked Questions
How do you find the sum of square numbers?
The sum of the first n squares, 1^{2}+2^{2}+...+n^{2} = n(n+1)(2n+1)/6.
For example, 1^{2}+2^{2}+...+10^{2}=10×11×21/6=385.
This result is usually proved by a method known as mathematical induction, and whereas it is a useful method for showing that a formula is true, it does not offer any insight into where the formula comes from. Instead we shall prove this result deductively, which will offer some insight into how you may add cubes, fourth powers, and so on.
Proof

r^{3}–(r–1)^{3}  =  n^{3}–(n–1)^{3} + (n–1)^{3}–(n–2)^{3} + ... + 3^{3}–2^{3} + 2^{3}–1^{3} + 1^{3}–0^{3}  
=  n^{3} 
But, r^{3}–(r–1)^{3} = r^{3} – (r^{3}–3r^{2}+3r–1) = 3r^{2}–3r+1.
∴ ∑ 3r^{2}–3r+1 = 3∑ r^{2} – 3∑ r + ∑1 = 3∑ r^{2} – 3n(n+1)/2 + n = n^{3}.
∴ 3∑ r^{2}  =  n^{3} + 3n(n+1)/2 – n 
=  2n^{3}+3n(n+1)–2n 2 

=  n(2n^{2}+3n+3–2) 2 

=  n(2n^{2}+3n+1) 2 

=  n(n+1)(2n+1) 2 

∴ ∑ r^{2}  =  n(n+1)(2n+1) 6 
To see how this deductive principle can be applied to higher powers, see the Sum of Cubes document.