Re: induction why is this wrong?

On 1 May, 07:34, chalong <chalon...@[EMAIL PROTECTED]
> wrote:
> Suppose that I am to prove by Inductive proof that F(n)G(n)=3D(2n)!,
n=991=

> and my F(n) is the (2k-1) from 1 to n and G(n) is 2k from 1 to n
> How do i prove this? I try this so far by i got stuck please help.
> P(n) =A0=3D =A0(2n-1)(2n)=3D(2n)! =A0 =A0 =A0 for n=991
> P(1) =A0 =A0=3D =A0 =A0 =A0 =A0(2*1-1)(2*1)=3D(2*1)!
> (1)(2)=3D(2*1)!
> 2=3D2 =A0TRUE
>
> Assume P(n) Prove P(n+1)
>
> Proof: =A0P(n+1) =A0=3D (2n-1)(2n)+(2(n+1)-1)*(2(n+1))=3D (2(n+1))!
> =3D (2n)!+2n+2n+2 =3D (2n+2)!
> =3D(2n)!+(4n+2)=3D(2n+2)!
> I cannot get that left hand side to equal (2n+2)! anyway =A0i tried.
> What is the correct way to do this?
> Also maybe i shouldn't replace F(n) by 2n-1 and G(n) by 2k by leave
> them in as part of proof?

You seem to have various characters/words missing from this.  I will
guess that your problem is:

Prove that F(n)*G(n)=3D(2n)! for n>=3D1
where F(n) is the product of (2k-1) from k=3D1 to k=3Dn
and G(n) is the product of (2k) from k=3D1 to k=3Dn.

Proof:
A) F(m+1)=3DF(m)*(2m+1)
B) G(m+1)=3DG(n)*(2m+2)
C) F(1)*G(1)=3D(2*1-1)*(2*1)=3D1*2=3D(2*1)! so hypothesis true for n=3D1
D) Assume hypothesis true for n=3Dm, then
F(m+1)*G(m+1)=3DF(m)*(2m+1)*G(m)*(2m+2)=3D(2m)!*(2m+1)*(2m+2)=3D(2m+2)!
so by induction true for integer n>=3D1