# Am I on the right track with this induction question?

1. Apr 26, 2004

### ptex

Step 0)
Code (Text):

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]

?

Last edited: Apr 26, 2004
2. Apr 26, 2004

### 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!
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.