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Math Help - sum prove

  1. #1
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    sum prove

    If U_n=\frac{1}{(2n+1)(2n+3)} , prove that

     <br />
U_{n-1}-U_n=\frac{4}{(2n-1)(2n+1)(2n+3)}<br />

    Thanks .
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  2. #2
    Air
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    Quote Originally Posted by thereddevils View Post
    If U_n=\frac{1}{(2n+1)(2n+3)} , prove that

     <br />
U_{n-1}-U_n=\frac{4}{(2n-1)(2n+1)(2n+3)}<br />

    Thanks .
    U_n = \frac{1}{(2n+1)(2n+3)}

    u_{n-1} is the point when considering a value one less than n thus the expression is found (for n-1) when we substitute n=n-1.

    U_{n-1} = \frac{1}{(2(n-1)+1)(2(n-1)+3)} = \frac{1}{(2n-2+1)(2n-2+3)} = \frac{1}{(2n-1)(2n+1)}

    \therefore U_{n-1} - U_n = \frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)}

    Bring the fraction into one by multiplying each fraction by common value.

    \frac{(2n+3) - (2n-1)}{(2n-1)(2n+1)(2n+3)}

    Simplify the numerator.

     \frac{4}{(2n-1)(2n+1)(2n+3)}
    Last edited by Air; February 18th 2009 at 02:25 AM. Reason: Grammer Error
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