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Math Help - Formula of a summation

  1. #1
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    Formula of a summation

    Hi all... I am having alot of trouble finding a guide on how to solve this. I have been told that I can compute the first few terms to get the formula but have no idea where to start. Thanks for your help in advance!
    Find a formula for 1/1*2 + 1/2*3 + 1/n*(n+1).
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by vexiked View Post
    Hi all... I am having alot of trouble finding a guide on how to solve this. I have been told that I can compute the first few terms to get the formula but have no idea where to start. Thanks for your help in advance!
    Find a formula for 1/1*2 + 1/2*3 + 1/n*(n+1).
    In summation form, this is \sum_{k=1}^n\frac{1}{k(k+1)}

    If you do a partial fraction decomposition on \frac{1}{k(k+1)}, you should end up with \frac{1}{k}-\frac{1}{k+1}

    Then, \sum_{k=1}^n\frac{1}{k(k+1)}=\sum_{k=1}^n\left[\frac{1}{k}-\frac{1}{k+1}\right]

    When you expand the sum, you end up with \sum_{k=1}^n\left[\frac{1}{k}-\frac{1}{k+1}\right]=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\frac{1}{n}-\frac{1}{n+1}. Now, a lot of cancellations occur, and we are left with 1-\frac{1}{n+1}

    So we can say that \sum_{k=1}^n\frac{1}{k(k+1)}=\sum_{k=1}^n\left[\frac{1}{k}-\frac{1}{k+1}\right]=1-\frac{1}{n+1}

    Does this make sense?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Since is...

    \frac{1}{k\cdot (k+1)}= \frac{1}{k}-\frac{1}{k+1}

    ... il also...

    \sum_{k=1}^{n} \frac{1}{k\cdot (k+1)}= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{n} - \frac{1}{n+1} = 1-\frac{1}{n+1}

    Kind regards

    \chi \sigma
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