first verify it holds for n=1
then you are allowed to assume it holds for n and you can start with the induction step
for n+1 instead of inserting it into the equation, add it to both sides and transform it to the original formula to proof its equality
[tex] 1^{2}+2^{2}+3^{2}+...+n^{2}+(n+1)^{2}=\frac{n(n+1)(2n+1)}{6} +(n+1)^{2} \\ =\frac{n(n+1)(2n+1)+6(n+1)^{2} }{6}\\ =\frac{n(n+1)(2n+1)+6(n+1)(n+1)}{6}\\ =\frac{(n+1)(n(2n+1)+6(n+1))}{6}\\ =\frac{(n+1)((2n^{2}+n)+6(n+1))}{6}\\ =\frac{(n+1)(2n^{2}+7n+6)}{6}\\ =\frac{(n+1)(n+2)(2n+3)}{6}\\ =\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}[/tex]
this equals the original formula if you had inserted n+1, proving the correctness
