# Mathematical induction?

• June 10th 2011, 07:17 PM
homeylova223
Mathematical induction?
Use mathematical induction to prove that each proposition is valid for all positive integral values of n

1^2+3^2+5^2+...+(2n-1)^2=n(2n-1)(2n+1)/3 This is what I did

(2k+1)^2

k(2k-1)(2k+1)/3

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

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

k(2k-1)(2k+1)/3 + 3(2k+1)(2k+1)/3 Now I am kind of stuck

Basically what I need is a formula that when n=k+1 is applied to n(2n-1)(2n+1)/3

it will give me the same result but I need to prove it.
• June 10th 2011, 07:38 PM
you've almost come to the end. All you have to do is simplify the final expression (where you said you are stuck) which eventually gives you $\frac{(k+1)(2k+1)(2k+3)}{3}$