Am I on the right track with this induction question?

[ptex]

Step 0)
22 + 42 + 62+...+(2n)2 = 2n(n+1)(2n+1)/3

Step 1)
Let n = 2
22 + 42
= 4 + 16
= 20

2(2)(2+1)(2(2)+1)/3
= 60/3
= 20

Step2)
Assume that the formula works for n=1,2,3,...,k
ie. 22 + 42 + 62+...+(2n)2 = 2k(k+1)(2k+1)/3
Step 3)
2n(n+1)(2n+1)/3 + (k+1)2

?

[arildno]

Step 1) is a correct verification of the formula.
Step 2) is very confusing, you seem to use n and k interchangeably!
n is the summation index, while the k'th term is the last to be summed!
Now instead:
Let 1<=n <= k:
Assume the proposition holds for the choice k.
We are to show that the proposition holds when summing 1<=n<=k+1:

Sum(from 1 to k+1)=Sum(from 1 to k)+(2(k+1))^(2)=

2k*(k+1)*(2k+1)/3+(2(k+1))^(2)

Now rearrange and try to gain the "formulaic prediction" for k+1.