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Math Help - Mathematical induction!

  1. #1
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    Mathematical induction!

    Hi,
    I could use some help with my proving. The assignment goes as follows:

    "Prove with mathematical induction that

    ((n+1)(n+2)...(2n)) / 22

    is a whole number with all values of n"

    I know the basic principle of induction but I can't see how can I show the "n+1" -step.

    Please excuse my inaccurate expressions, my native language is finnish I would REALLY appreciate the help!
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  2. #2
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    Re: Mathematical induction!

    Do you mean that (n+1)(n+2)...(2n) is divisible by 4? This is obvious because 2n is even and, since the number of factors is ≥ 3 for n ≥ 3, there is another even factor.
    Last edited by emakarov; November 15th 2012 at 12:20 PM. Reason: Changed "≥ 1" to "≥ 3".
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  3. #3
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    Re: Mathematical induction!

    What, exactly, do you mean by ((n+1)(n+ 2)...(2n)) if n= 1? I see that, for example, if n= 3, then this is 4(5)(6) but what about n= 1? Is this "2" or "2(2)"? If the first, which is what I would think because 1+ 1= 2(1), the statement is not true: (1+ 1)/ 2^2= 1/2

    But for n= 2, this is 3(4)(4) which is divisible by 2^2= 4 because it has a factor of 4. If n\ge 4, there are four consecutive numbers in that so at least one is divisible by 4.
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  4. #4
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    Re: Mathematical induction!

    Okay so here is a hopefully a bit better presentation of the assignment:

    "Prove with mathematical induction that

     \frac{(n+1)(n+2)...(2n)}{2^n}

    is an integer when  n \in  \mathbb{N}

    First time using latex...

    thank you for your help!
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  5. #5
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    Re: Mathematical induction!

    ... and theres minor error in my first post. sorry bout that
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Mathematical induction!

    I chose to write the induction hypothesis P_n after looking at the first several statements:

    \frac{(2n)!}{n!}=(2n-1)!!2^n

    We easily see that P_1 is true, so next I defined:

    \mu(n)=\frac{(2(n+1))!}{(n+1)!}-\frac{(2n)!}{n!}=\frac{(2(n+1))!-(n+1)(2n)!}{(n+1)!}=\frac{(2n)!((2n+2)(2n+1)-(n+1))}{(n+1)!}=

    \frac{(2n)!}{n!}\(2(2n+1)-1\)=\frac{(2n)!}{n!}(4n+1)=(2n-1)!!2^n(4n+1)

    Now, adding \mu(n) to both sides of P_n there results:

    \frac{(2(n+1))!}{(n+1)!}=(2n-1)!!2^n+(2n-1)!!2^n(4n+1)

    \frac{(2(n+1))!}{(n+1)!}=(2n-1)!!2^n(4n+2)

    \frac{(2(n+1))!}{(n+1)!}=(2n-1)!!2^{n+1}(2(n+1)-1)

    \frac{(2(n+1))!}{(n+1)!}=(2(n+1)-1)!!2^{n+1}

    We have derived P_{n+1} from P_n thereby completing the proof by induction.
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