This is like the 1000th induction promblem i posted but im trying to get the hang of this.

Use mathematical induction to prove: F(n)G(n) = (2n)!, n>=1

F(n)= 1*3*5...*(2n-1)

G(n) = 2k = 2*4*6*... *2n

1. P(n): 1*3*5*...*(2n-1)*2*4*6*...*2n = (2n)!

2. P(1): 1*3*5*...*(2(1)-1)*2*4*6*...*2(1) = (2*1)!

2=2! True

3. Assume: 1*3*5*...*(2n-1)*2*4*6*...*2n = (2n)!

4. Prove P(n+1)

1*3*5*...*(2n-1)*2*4*6*...*2n = (2n)!

1*3*5*...*(2n-1)*2*4*6*2(n+1) = (2n)!

QED ((2n)!*2(n+1))= (2n+1)!

This is true when i plug in one, but when i plugged in higher numbers they weren't equal. spent a lot of time trying to figure out where the fallacy was. Any insights would be appreciated, thanks!