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Math Help - Cant find my error Mathematical Induction

  1. #1
    Junior Member
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    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!
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  2. #2
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    Hello, tokio!

    You almost had it . . .


    \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.

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