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Math Help - Simplifying Expression with Like Denominators and Factorials

  1. #1
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    Simplifying Expression with Like Denominators and Factorials

    Hi,

    I am having difficulties simplifying the expression  \frac{1}{k!}-\frac{1}{(k+1)!} .

    To obtain like denominators,  \frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)!}{k!(k+1)!}
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  2. #2
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    Re: Simplifying Expression with Like Denominators and Factorials

    Since (k+1)! = (k+1)k! = (k+1)k(k-1)!,

    \frac{1}{k!}-\frac{1}{(k+1)!} = \frac{1}{k!}\left(1-\frac{1}{k+1}\right)=\frac{1}{k!}\cdot\frac{k}{k+1  } =\frac{1}{(k-1)!(k+1)}
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  3. #3
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    Re: Simplifying Expression with Like Denominators and Factorials

    Hi,

    The answer according to Maple is  \frac{k}{(k+1)!} .

    Is it wrong?
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  4. #4
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    Re: Simplifying Expression with Like Denominators and Factorials

    Quote Originally Posted by SC313 View Post
    Is it wrong?
    No, both are correct. You should review the definition of factorial.
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  5. #5
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    Re: Simplifying Expression with Like Denominators and Factorials

    Could you please explain? I still am confused.
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    Re: Simplifying Expression with Like Denominators and Factorials

    Have you reviewed and understood the definition of factorial?
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    Re: Simplifying Expression with Like Denominators and Factorials

    Yes, n! is defined for a positive integer n as n!=n(n-1)...2*1.

    Is that not what you mean?
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    Re: Simplifying Expression with Like Denominators and Factorials

    Yes, that what I meant. So, if you understand the definition, how can you not see that

    \frac{k}{(k+1)!}=\frac{k}{(k+1)k(k-1)\cdots1} =(\text{cancel }k)\frac{1}{(k+1)(k-1)\cdots1} =\frac{1}{(k+1)(k-1)!}?
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    Re: Simplifying Expression with Like Denominators and Factorials

    Quote Originally Posted by SC313 View Post
    the expression  \frac{1}{k!}-\frac{1}{(k+1)!} .
     \frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)}{(k+1)(k!)}-\frac{1}{(k+1)!}=\frac{k}{(k+1)!}
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  10. #10
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    Re: Simplifying Expression with Like Denominators and Factorials

    Quote Originally Posted by Plato View Post
     \frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)}{(k+1)(k!)}-\frac{1}{(k+1)!}=\frac{k}{(k+1)!}
    Do you not need like denominators? Should both denominators not be  (k+1)k! ?
    It seems it would be  \frac{(k+1)}{(k+1)(k!)}-\frac{k!}{(k+1)!k!}
    I just really am frustrated and confused.
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  11. #11
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    Re: Simplifying Expression with Like Denominators and Factorials

    (k+1)k! = (k+1)k(k-1)\cdots 2\cdot 1 = (k+1)!

    Example: 5! = 5\cdot 4! = 5\cdot 4\cdot 3! = 5\cdot 4\cdot 3\cdot 2! = 5\cdot 4\cdot 3\cdot 2\cdot 1!

    Edit: In other words, they already had like denominators, but

    (k+1)!k! = [(k+1)k(k-1)\cdots 2\cdot 1][k(k-1)\cdots 2\cdot 1] = (k+1)k^2(k-1)^2\cdots 2^2\cdot 1^2

    So, if you multiply (k+1)! by k!, you don't get (k+1)k!.
    Last edited by SlipEternal; November 1st 2013 at 08:09 PM.
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    Re: Simplifying Expression with Like Denominators and Factorials

    Quote Originally Posted by SC313 View Post
    Hi,

    I am having difficulties simplifying the expression  \frac{1}{k!}-\frac{1}{(k+1)!} .

    To obtain like denominators,  \frac{1}{k!}-\frac{1}{(k+1)!}=\frac{(k+1)!}{k!(k+1)!}
    Notice that \displaystyle \begin{align*} \left( k + 1 \right) ! = \left( k + 1 \right) k! \end{align*}, so the lowest common denominator is \displaystyle \begin{align*} \left( k + 1 \right) k! \end{align*}, or just \displaystyle \begin{align*} \left( k + 1 \right) ! \end{align*}. So

    \displaystyle \begin{align*} \frac{1}{k!} - \frac{1}{\left( k + 1 \right) !} &= \frac{k + 1}{\left( k + 1 \right) k!} - \frac{1}{ \left( k + 1 \right) ! } \\ &= \frac{k + 1}{ \left( k + 1 \right) ! } - \frac{1}{ \left( k + 1 \right) ! } \\ &= \frac{k}{ \left( k + 1 \right) ! } \end{align*}
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