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Math Help - sum of telescoping series...

  1. #1
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    Exclamation sum of telescoping series...

    i need to find the sum of :



    I'm getting the wrong answer, just cant find out where i'm going wrong.

    here's what i did...

    replaced the above by;















    & the given answer is 35/429
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  2. #2
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    Quote Originally Posted by adhyeta View Post
    i need to find the sum of :



    I'm getting the wrong answer, just cant find out where i'm going wrong.

    here's what i did...

    replaced the above by;















    & the given answer is 35/429
    you could split it up further

     <br />
\frac{1}{(2n-1)(2n+1)(2n+3)} = \frac{1}{8} \left(<br />
\frac{1}{2n-1} - \frac{1}{2n+1} + \frac{1}{2n+3}\right)<br />
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  3. #3
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    Hello, adhyeta!

    Find the sum: . S \;=\;\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot  7}+\frac{1}{5\cdot7\cdot9}+\frac{1}{7\cdot9\cdot11  }+\frac{1}{9\cdot11\cdot13}

    The given answer is: \frac{35}{429}
    I would make three fractions: . \frac{1}{(2n-1)(2n+1)(2n+3)} \;=\;\frac{1}{8}\,\bigg[\frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3}\bigg]


    Then we have:

    . . \frac{1}{1\cdot3\cdot5} \;=\; \frac{1}{8}\left[\frac{1}{1} - \frac{2}{3} + \frac{{\color{cyan}\rlap{/}}1}{{\color{cyan}\rlap{/}}5}\right]

    . . \frac{1}{3\cdot5\cdot7} \;=\;\frac{1}{8}\bigg[\frac{1}{3} - \frac{{\color{cyan}\rlap{/}}2}{{\color{cyan}\rlap{/}}5} + \frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7}\bigg]

    . . \frac{1}{5\cdot7\cdot9} \;=\;\frac{1}{8}\bigg[\frac{{\color{cyan}\rlap{/}}1}{{\color{cyan}\rlap{/}}5} - \frac{{\color{red}\rlap{/}}2}{{\color{red}\rlap{/}}7} + \frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9}\bigg]

    . . \frac{1}{7\cdot9\cdot11} \;=\;\frac{1}{8}\bigg[\frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7} - \frac{{\color{green}\rlap{/}}2}{{\color{green}\rlap{/}}9} + \frac{1}{11}\bigg]

    . \frac{1}{9\cdot11\cdot13} \;=\;\frac{1}{8}\bigg[\frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9} - \frac{2}{11} + \frac{1}{13}\bigg]



    Add the equations:

    . . S \;=\;\frac{1}{8}\bigg[1 - \frac{2}{3} + \frac{1}{3} + \frac{1}{11} - \frac{2}{11} + \frac{1}{13}\bigg] \;=\;\frac{1}{8}\cdot \frac{280}{429} \;=\;\boxed{\frac{35}{429}}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, adhyeta!

    I would make three fractions: . \frac{1}{(2n-1)(2n+1)(2n+3)} \;=\;\frac{1}{8}\,\bigg[\frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3}\bigg]


    Then we have:

    . . \frac{1}{1\cdot3\cdot5} \;=\; \frac{1}{8}\left[\frac{1}{1} - \frac{2}{3} + \frac{{\color{cyan}\rlap{/}}1}{{\color{cyan}\rlap{/}}5}\right]

    . . \frac{1}{3\cdot5\cdot7} \;=\;\frac{1}{8}\bigg[\frac{1}{3} - \frac{{\color{cyan}\rlap{/}}2}{{\color{cyan}\rlap{/}}5} + \frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7}\bigg]

    . . \frac{1}{5\cdot7\cdot9} \;=\;\frac{1}{8}\bigg[\frac{{\color{cyan}\rlap{/}}1}{{\color{cyan}\rlap{/}}5} - \frac{{\color{red}\rlap{/}}2}{{\color{red}\rlap{/}}7} + \frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9}\bigg]

    . . \frac{1}{7\cdot9\cdot11} \;=\;\frac{1}{8}\bigg[\frac{{\color{red}\rlap{/}}1}{{\color{red}\rlap{/}}7} - \frac{{\color{green}\rlap{/}}2}{{\color{green}\rlap{/}}9} + \frac{1}{11}\bigg]

    . \frac{1}{9\cdot11\cdot13} \;=\;\frac{1}{8}\bigg[\frac{{\color{green}\rlap{/}}1}{{\color{green}\rlap{/}}9} - \frac{2}{11} + \frac{1}{13}\bigg]



    Add the equations:

    . . S <br /> <br />
\;=\;\frac{1}{8}\bigg[1 - \frac{2}{3} + \frac{1}{3} + \frac{1}{11} - \frac{2}{11} + \frac{1}{13}\bigg] \;=\;\frac{1}{8}\cdot \frac{280}{429} \;=\;\boxed{\frac{35}{429}}


    hey soroban. i got that.
    thnx

    but what do you think might be wrong with my solution.
    i have four terms you have combined 2 of them as one & then solved.

    but doing it the other way should also give the same answer, right?
    what do you think might be wrong with it?
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  5. #5
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    we have \frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{4}\left( \frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)} \right), thus from here we've automatically found a telescoping sum, thus your sum equals \frac{1}{4}\left( \frac{1}{3}-\frac{1}{11\times 13} \right)=\frac{1}{4}\times \frac{140}{429}=\frac{35}{429}.
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  6. #6
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    Hello, adhyeta!

    Find the sum:

    \frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\f  rac{1}{5\cdot7\cdot9}+\frac{1}{7\cdot9\cdot11}+\fr  ac{1}{9\cdot1\cdot.13}

    I'm getting the wrong answer, just cant find out where i'm going wrong.

    here's what i did...

    replaced the above by;

    \sum_{n=1}^{5}\frac{1}{(2n-1)(2n+1)(2n+3)}

    . . =\; \frac{1}{2}\sum_{n=1}^{5}\frac{(2n+3)-(2n+1)}{(2n-1)(2n+1)(2n+3)}

    . . =\; \frac{1}{2}\sum_{n=1}^{5}\left[\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n-1)(2n+3)}\right]

    . . =\; \frac{1}{2}\sum_{n=1}^{5}\left[\frac{1}{2}\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}-\frac{1}{4}\frac{(2n+3)-(2n-1)}{(2n-1)(2n+3)}\right]

    . . =\;\frac{1}{4}\sum_{n=1}^{5}\left[\frac{1}{(2n-1)}-\frac{1}{(2n+1)}-\frac{1}{2}\frac{1}{(2n-1)}+\frac{1}{2}\frac{1}{(2n+3)}\right]
    At this point, I'd factor again . . .

    \frac{1}{8}\sum^5_{n=1}\left[\frac{2}{2n-1} - \frac{2}{2n+1} - \frac{1}{2n-1} + \frac{1}{2n+3}\right]

    . . = \; \frac{1}{8}\sum^5_{n=1}\bigg[\frac{1}{2n-1} - \frac{2}{2n+1} + \frac{1}{2n+3}\bigg]


    And this is equivalent to what I had found . . .

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