Thread: Cant find my error Mathematical Induction

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!

Hello, tokio!

You almost had it . . .

Use mathematical induction to prove: .

Here's what I would do . . .

Multiply both sides by the next odd number, ,
and multiply both sides by the next even number, .

. .

. .


We have: .


We have proved . . . The inductive proof is compete.