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Math Help - Method of telescopic sums....help

  1. #1
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    Method of telescopic sums....help

    hi all i need to solve,

    Using method of telescopics sums to determine that;

    999
    Σ 2k+1/k^2(k+1)^2 = 0.000099
    k=100


    Please help!!!!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Craige View Post
    hi all i need to solve,

    Using method of telescopics sums to determine that;

    999
    Σ 2k+1/k^2(k+1)^2 = 0.000099
    k=100
    As you did not use enough parenthesis, I guess we should understand \sum_{100}^{999}\frac{2k+1}{k^2(k+1)^2}. Then note that \frac{1}{k^2}-\frac{1}{(k+1)^2}=\frac{2k+1}{k^2(k+1)^2}. (check it !) Using this, you should be able to conclude. Good luck !
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by flyingsquirrel View Post

    Then note that \frac{1}{k^2}-\frac{1}{(k+1)^2}=\frac{2k+1}{k^2(k+1)^2}. (check it !)
    The thing is, not all people is able to see this. Of course, pretty easy when given the partial fraction descomposition and then just do algebra to verify it.

    To make what flyingsquirrel did, first note that 2k+1=(k^2+2k+1)-k^2, nothing weird, we just added & subtracted a convenient term. Following up on this, we have that 2k+1=(k+1)^2-k^2. Now, what's the cool thing here: the cool here tell us that in the denominator we have exactly the terms k^2 & (k+1)^2. Hence, \frac{2k+1}{k^2(k+1)^2}=\frac{(k+1)^2-k^2}{k^2(k+1)^2}.

    Finally, split this original ratio into two parts, like \frac{x+y}{z}=\frac xz+\frac yz, and the last ratio becomes \frac{(k+1)^2}{k^2(k+1)^2}-\frac{k^2}{k^2(k+1)^2}. Then after makin' simple simplifications, we happily get that \frac{2k+1}{k^2(k+1)^2}=\frac1{k^2}-\frac1{(k+1)^2}.

    -----

    Back to the sum and let a_k=\frac1{k^2}\implies a_{k+1}=\frac1{(k+1)^2}. So we can see the sum as \sum_{k=100}^{900}(a_k-a_{k+1}). Add some terms: (a_{100}-a_{101})+(a_{101}-a_{102}) and so on. Can you see what's goin' on here?
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