Re: induction why is this wrong?

> if you want to show that F(n) < G(n), why don't you put it in?
> Assuming you have the same F(n) and G(n), then take G(n)-F(n)=3D
> \sum_1^n (2k) -\sum_1^n(2k-1)=3D\sum_1^n(2k-(2k-1)=3D\sum_1^n(1)=3Dn>0,
th=
us
> result. Or is F(n)(G(n)) the product from 1 to n of 2k-1 (2k)?

F(n) the product of the first n odd positive integers
G(n) the product of the first n even positive integers

Thus I plug in and did this:
4.Inductive proof that F(n)<G(n),	 n=991
P(n): F(n)<G(n)
P(1): F(1)<G(1)=3D(2*1-1)<(2*1)
=3D1<2 True
Assume P(n), that F(n)<G(n) Prove for P(n+1)
Proof:	F(n+1)=3DF(n)*(2n+1)
=3D(2n-1)(2n+1)
=3D 4n^2- 4n-1<G(n)(2n+2)
=3D 4n^2- 4n-1<(2n)(2n+2)
=3D4n^2- 4n-1<4n^2+2n

Do you think that this is valid?