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Math Help - sequences 1

  1. #1
    Senior Member furor celtica's Avatar
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    sequences 1

    hey i'm truly sorry if this isnt in the right forum, i actually couldnt find a subforum that addressed sequence problems
    so this is a basic problem inolving factorials so don't make the answer too complicated please

    show that (2n)!/n! = 2^n (1x3x5...x(2n -1))

    i really have very little work to show cos i got nowhere in the hours i've been trying to wrap my head around this. its probably simple and i'm overestimating it but the farthest i've got is to 2n x (2n-1) x (2n-2) x (2n-3) x...x (n)
    i'm not even sure that is correct
    any help would be appreciated, please show all working and details, i'd like to totally understand this
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  2. #2
    Moo
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    Hello,

    In (2n)! you have 2n, 2n-2, 2n-4,... and 1,3,5,...,2n-1
    Now what if you factor out 2 of each of 2n,2n-2,... ?
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  3. #3
    Senior Member furor celtica's Avatar
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    um, this is not so clear, what do you mean by
    'you have 2n, 2n-2, 2n-4,... and 1,3,5,...,2n-1'
    ?
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  4. #4
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    Hello, furor celtica!

    \text{Show that: }\;\dfrac{(2n)!}{n!} \:=\: 2^n(1\cdot3\cdot 5 \cdots [2n -1])

    Did you examine a specific case?


    Let n = 4.

    Then we have: . \dfrac{8!}{4!} \;=\;\dfrac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\  cdot1}{4\cdot3\cdot2\cdot1}


    Reduce: . \dfrac{\not8^{^2}\! \cdot 7 \;\cdot\!\not6^{^2}\! \cdot5\; \cdot \!\not4^{^2}\! \cdot3 \;\cdot \!\not2^{^2}\cdot 1}{\not4\; \cdot\! \not3 \;\cdot\! \not2 \;\cdot\! \not1}


    . . and we have: . 2^4(1\cdot3\cdot5\cdot7)



    Do you see what happened?

    The denominator cancels into the even factors in the numerator
    . . leaving behind a "bunch of 2's" ... and all the odd factors.


    Now can you express all that in general terms?

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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by furor celtica View Post
    um, this is not so clear, what do you mean by
    'you have 2n, 2n-2, 2n-4,... and 1,3,5,...,2n-1'
    ?
    You separate the (2n)! factorial into the product of odd and even terms

    CB
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  6. #6
    Senior Member furor celtica's Avatar
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    ok so i started with the RHS and got to ((2^n)(2n!))/(2x4x6x...x2n), i.e. for the statement to be true i have to prove that (2x4x6x...x2n) = (2^n)(n!)
    does this need to be proven really or is it obvious?
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