Prove for all n∈N $1^2+3^2+5^2+...+(2n-1)^2=\frac{4n^3-n}{3}$ 
Prove for all n∈N $1^2+3^2+5^2+...+(2n-1)^2=\frac{4n^3-n}{3}$
  This is what I got, but I dont think it is right.


The fact that you are dealing with a sum of squares seems to have completely disappeared in the step.
Here's what you need:
$$1^2+3^2+...+(2k-1)^2+(2k+1)^2= \text{ (Inductive Hypothesis)}$$
$$\frac{4k^3-k}{3}+(2k+1)^2= $$
$$\frac{4k^3-k}{3}+\frac{3(2k+1)^2}{3}= $$
$$\frac{4k^3-k+3(2k+1)^2}{3}= $$
$$\frac{4k^3-k+3(4k^2+4k+1)}{3}= $$
$$\frac{4k^3-k+12k^2+12k+3}{3}= $$
$$\frac{4k^3+12k^2+12k+3-k+1-1}{3}= $$
$$\frac{4k^3+12k^2+12k+4-(k+1)}{3}$$
$$\frac{4(k^3+3k^2+3k+1)-(k+1)}{3}$$
$$\frac{4(k+1)^3-(k+1)}{3}$$
