
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+...+(2n1)^2=n(2n1)(2n+1)/3 This is what I did
(2k+1)^2
k(2k1)(2k+1)/3
k(2k1)(2k+1)/3 + (2k+1)^2/1
k(2k1)(2k+1)/3 + 3(2k+1)^2/3
k(2k1)(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(2n1)(2n+1)/3
it will give me the same result but I need to prove it.

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 $\displaystyle \frac{(k+1)(2k+1)(2k+3)}{3}$

How would you simplify it do you just use distributive property because the (2k+1) on the left side is confusing me.

why, (2k+1) is common to both sides, take it out and simplify