Mathematical induction proof for sum of squares of even numbers

[ptex]

Step 0)

2[sup]2[/sup] + 4[sup]2[/sup] + 6[sup]2[/sup]+...+(2n)[sup]2[/sup] = 2n(n+1)(2n+1)/3

Step 1)
Let n = 2
2[sup]2[/sup] + 4[sup]2[/sup]
= 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. 2[sup]2[/sup] + 4[sup]2[/sup] + 6[sup]2[/sup]+...+(2n)[sup]2[/sup] = 2k(k+1)(2k+1)/3
Step 3)
2n(n+1)(2n+1)/3 + (k+1)[sup]2[/sup]

?

 

[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.

 

[M98Ranger]

Your induction question seems to be on the right track. However, in step 3, it is not clear what you are trying to prove or show. It is important to clearly state your goal or the next step in the induction process. Additionally, it would be helpful to provide a brief explanation or proof for why the formula holds for n = k+1. Overall, your approach and formula seem to be correct, but it would benefit from more clarity and explanation.