# Cant find my error Mathematical Induction

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• April 21st 2009, 11:35 AM
tokio
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!
• April 21st 2009, 01:00 PM
Soroban
Hello, tokio!

You almost had it . . .

Quote:

$\begin{array}{ccc}F(n) &=& 1 \cdot 3 \cdot 5\hdots (2n-1) \\ G(n) &=& 2\cdot4\cdot6\cdots 2n \end{array}$

Use mathematical induction to prove: . $F(n)\!\cdot\!G(n) \:=\: (2n)!$

$1.\;\;P(n)\!:\;\bigg[1\cdot3\cdot5 \cdots(2n-1)\bigg]\,\bigg[2\cdot4\cdot6\cdots 2n\bigg] \;=\; (2n)!$

$2.\;\;P(1)\!:\;\;1\cdot2 \:=\: (2\!\cdot\!1)! \quad\hdots \text{true}$

$3.\;\;\text{Assume }S(k)\!:\;\bigg[1\cdot3\cdot5\cdots(2k-1)\bigg]\,\bigg[2\cdot4\cdot6\cdots2k\bigg] \;=\; (2k)!$

Here's what I would do . . .

Multiply both sides by the next odd number, $2k+1$,
and multiply both sides by the next even number, $2k+2$.

. . $\bigg[1\ \cdot\ 3\ \cdot\ 5\ \cdots(2k-1)\bigg]\,\bigg[2\ \cdot\ 4\ \cdot\ 6\ \cdots 2n\bigg]\cdot{\color{blue}(2k+1)(2k+2)} \;=\;(2k)!\cdot{\color{blue}(2k+1)(2k+2)}$

. . $\bigg[1\cdot3\cdot5\cdots(2k+1)\bigg]\,\bigg[2\cdot4\cdot6\cdots(2n+2)\bigg] \;=\;\underbrace{(2n)!\cdot(2n+1)\cdot(2n+2)}_{\te xt{This is }(2n+2)!}$

We have: . $\bigg[1\cdot2\cdot3\cdots(2k+1)\bigg]\,\bigg[2\cdot4\cdot5\cdots2(k+1)\bigg] \;=\;\bigg[2(k+1)\bigg]!$

We have proved $S(k+1)$ . . . The inductive proof is compete.