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Thread: [SOLVED] induction integer question

  1. #1
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    [SOLVED] induction integer question

    Prove that for any integer n ≥ 1,
    (2n)!/2^n is an integer!

    Well I started off saying let n=(2n!)/2^n (*)

    and did the usual step like let n=1 (which is obviously true)

    then when I go to let n=n+1

    i get

    n + (n+1) = (2(n+1))!/2^(n+1)

    i can substitute the LHS n with (*)

    but then I get stuck and dont know what to do

    so im guessing my method is incorrect :\

    can someone shed some light on this.. thanks!
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  2. #2
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    Hello, Khonics89!

    Prove that for any integer $\displaystyle n \geq 1,\;\frac{(2n)!}{2^n}$ is an integer.
    You started correctly . . .

    $\displaystyle S(n)\!:\;\;\frac{(2n)!}{2^n} \:=\:a,\:\text{ for some integer }a.$


    Verify $\displaystyle S(1)\!:\;\;\frac{2!}{2^1} \:=\:1$ . . . True.

    Assume $\displaystyle S(k)$ is true: . $\displaystyle \frac{(2k)!}{2^k} \:=\:a,\:\text{ for some integer }a.$


    We want to prove that $\displaystyle S(k+1)$ is true: . $\displaystyle \frac{(2[k+1])!}{2^{k+1}}$ is an integer.


    We have: .$\displaystyle \frac{(2[k+1])!}{2^{k+1}} \;\;=\;\;\frac{(2k+2)!}{2^{k+1}} \;=\;\frac{(2k+2)(2k+1)(2k)!}{2\cdot2^k} $

    . . $\displaystyle = \;\;\frac{(2k+2)(2k+1)}{2}\cdot\frac{(2k)!}{2^k} \;\;=\;\;\frac{2(k+1)(2k+1)}{2}\cdot\frac{(2k)!}{2 ^k}$

    . . $\displaystyle =\;\;(k+1)(2k+1)\cdot\underbrace{\frac{(2k)!}{2^k} }_{\text{This is }a} \;\;=\;\;\underbrace{(k+1)(2k+1)\!\cdot\!a }_{\text{This is an integer}}$


    We have proved $\displaystyle S(k+1)$ . . . The inductive proof is complete.

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  3. #3
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    Thanks- yeah I never saw it like that basicaly I did the proof but didn't know it was right .. soo yeah thanks once again
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