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Math Help - simplifying n factorials

  1. #1
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    simplifying n factorials

    can someone explain this process:

    the original question is f(n) = [(2^n)(n!)]/(n+2)!] and the solution says it simplifies to f(n) = [(2^n)]/ (k+1)(k+2)] can someone explain to me how the cancelled out the factorials and the addition of the (k+1) in the denominator?

    thanks
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  2. #2
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    not sure about the k part but

     f(n) = \frac{2^nn!}{(n+2)!} = \frac{2^nn!}{(n+2)(n+1)n!}= \frac{2^n}{(n+2)(n+1)}<br />
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  3. #3
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    Quote Originally Posted by pickslides View Post
    not sure about the k part but

     f(n) = \frac{2^nn!}{(n+2)!} = \frac{2^nn!}{(n+2)(n+1)n!}= \frac{2^n}{(n+2)(n+1)}<br />
    wait for the 2nd step how did u get (n+1)n!??
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  4. #4
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    Quote Originally Posted by johntuan View Post
    wait for the 2nd step how did u get (n+1)n!??
    (n+2)! = (n+2)(n+1)\textcolor{red}{(n)(n-1)(n-2) ...}

    what is that in red?
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  5. #5
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    Re: simplifying n factorials

    1 + [tex] cos [tex] theta [tex] / 2
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  6. #6
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    Re: simplifying n factorials

    Quote Originally Posted by anil86 View Post
    1 + [tex] cos [tex] theta [tex] / 2
    What on Earth?? Did you mean this to go into another thread or something?

    -Dan
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  7. #7
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    Re: simplifying n factorials

    Quote Originally Posted by johntuan View Post
    wait for the 2nd step how did u get (n+1)n!??
    It comes directly from the definition of "factorial". n! is defined as "n(n-1)(n-2)...(3)(2)(1)". That is "n times all positive integers less than n". (n+2)! is "n+ 2 times all positive integers less than n+ 2" or (n+2)(n+1)(n)(n-1(n-2)...(3)(2)(1)= (n+2)(n+1)(n!).
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