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.

 A: The fact that you are dealing with a sum of squares seems to have completely disappeared in the step.
Here's what you need:
$$1^2+3^2+...+(2k-1)^2+(2k+1)^2= \text{ (Inductive Hypothesis)}$$
$$\frac{4k^3-k}{3}+(2k+1)^2= $$
$$\frac{4k^3-k}{3}+\frac{3(2k+1)^2}{3}= $$
$$\frac{4k^3-k+3(2k+1)^2}{3}= $$
$$\frac{4k^3-k+3(4k^2+4k+1)}{3}= $$
$$\frac{4k^3-k+12k^2+12k+3}{3}= $$
$$\frac{4k^3+12k^2+12k+3-k+1-1}{3}= $$
$$\frac{4k^3+12k^2+12k+4-(k+1)}{3}$$
$$\frac{4(k^3+3k^2+3k+1)-(k+1)}{3}$$
$$\frac{4(k+1)^3-(k+1)}{3}$$
A: $\sum_{k=1}^n (2k-1) = \frac{4n^3-n}{3}$ is what you want to prove for all n. So I'm skipping the base case and moving to the inductive step
In the inductive step you supposed the formula is true for n, and try to prove it for n + 1
so now what we want to prove is $$\sum_{k=1}^{n+1} (2k-1) = \frac{4(n+1)^3-(n+1)}{3}$$
For summation, these cases are often quite easy for simple induction. remove the last term from the summation (that is, the "(n+1)th" term) and add it normally
you'll end up with
$$\sum_{k=1}^{n+1} (2k-1)= \sum_{k=1}^n (2k-1) + [2(n+1)-1] $$
But because of our assumption, we know that the summation on the right equals our formula, so 
$$\sum_{k=1}^n (2k-1) + [2(n+1)-1] = \frac{4n^3-n}{3} + [2(n+1)-1] $$
The rest is just ordering everything to try and reach the original formula
