Thread: 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.

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

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