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Double Factorial

  1. Oct 14, 2011 #1
    How to prove

    [tex](2n)!!=2^nn![/tex]

    for example

    [tex](2)!!=2^11!=2[/tex]

    [tex](4)!!=2^22!=8[/tex]

    ...

    [tex](2n)!!=2n(2n-2)(2n-4)...=2^3n(n-1)(n-2)...[/tex]

    I see that by intiution but I don't know how to write prove.
     
  2. jcsd
  3. Oct 14, 2011 #2

    Mute

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    Prove it by induction. You've shown that it holds for the case n =1, so if you assume that it holds for general n, show that it follows that it holds for the case n+1.
     
  4. Oct 14, 2011 #3

    dextercioby

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    You don't need induction. A direct computation using the definition of the double factorial is enough.
     
  5. Oct 14, 2011 #4
  6. Oct 14, 2011 #5
    [tex]
    (2 n)!! = 2 \times 4 \times \ldots \times (2 n)
    [/tex]

    But, notice that:
    [tex]
    \begin{array}{l}
    2 = 2 \times 1 \\

    4 = 2 \times 2 \\

    \ldots \\

    2 n = 2 \times n
    \end{array}
    [/tex]

    Combine the first factor of two from each factor in the double factoriel. How many of them are there? What do the remaining factors give?
     
  7. Oct 14, 2011 #6
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