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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

    Express 1/r(r+1) in partial fractions and use the result to find the sum of the following series as a single fraction in terms of n:

    1/1x2 + 1/2x3 + 1/3x2 + ............... + 1/n(n+1)

    I can express it as partial fractions to 1/r - 1/r-1 but from there, I don't know, I don't think it's an arithmetic sequence as there is no common difference.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Leodinas View Post
    Express 1/r(r+1) in partial fractions and use the result to find the sum of the following series as a single fraction in terms of n:

    1/1x2 + 1/2x3 + 1/3x2 + ............... + 1/n(n+1)

    I can express it as partial fractions to 1/r - 1/r-1 but from there, I don't know, I don't think it's an arithmetic sequence as there is no common difference.

    Thanks in advance.
    The partial fraction decomposition is \frac{1}{r} - \frac{1}{r{\color{red}+}1}. Then the series can be written:

    \left(\frac{1}{1} - \frac{1}{1+1}\right) + \left(\frac{1}{2} - \frac{1}{2+1}\right) + \left(\frac{1}{3} - \frac{1}{3+1}\right) + \, .... \, + \left(\frac{1}{n} - \frac{1}{n+1}\right)


    = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \, .... \, + \left(\frac{1}{n} - \frac{1}{n+1}\right)


    and the rest should be clear.
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  3. #3
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    Thank you for your help, I still don't see where to go with it after that, could you give me another clue?

    Similarly,

    \frac{a^2-b^2}{(ax+b)(bx+a)}

    I end up with two seemingly unhelpful simultaneous equations, any advice or direction for this question would be appreciated also.
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  4. #4
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    Quote Originally Posted by Leodinas View Post
    Thank you for your help, I still don't see where to go with it after that, could you give me another clue? [snip]
    Perhaps the answer will make it clear what you have to do: 1 - \frac{1}{n+1}.

    Quote Originally Posted by Leodinas View Post
    [snip]
    \frac{a^2-b^2}{(ax+b)(bx+a)}

    I end up with two seemingly unhelpful simultaneous equations, any advice or direction for this question would be appreciated also.
    It will be easier to help if you show all of the work you've done.

    For your future reference: Ask new questions in a new thread.
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  5. #5
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    \frac{a^2-b^2}{(ax+b)(bx+a)} = \frac{A}{ax+b} + \frac{B}{bx+a}

    a^2-b^2 = A(bx+a) + B(ax+b)

    Comparing x: 0 = Ab + Ba

    Comparing constants: a^2-b^2 = Aa + Bb

    I also recognise that a^2-b^2 can be (a+b)(a-b).
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  6. #6
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    Quote Originally Posted by Leodinas View Post
    \frac{a^2-b^2}{(ax+b)(bx+a)} = \frac{A}{ax+b} + \frac{B}{bx+a}

    a^2-b^2 = A(bx+a) + B(ax+b)

    Comparing x: 0 = Ab + Ba

    Comparing constants: a^2-b^2 = Aa + Bb

    I also recognise that a^2-b^2 can be (a+b)(a-b).
    0 = Ab + Ba .... (1)

    a^2-b^2 = Aa + Bb .... (2)

    Multiply equation (1) by -a and equation (2) by b. Add the results:

    b(a^2 - b^2) = b^2 B - a^2 B \Rightarrow b(a^2 - b^2) = -B (a^2 - b^2) \Rightarrow B = -b.

    Use this value to solve for A.
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    Thank you for that, I never would have thought to do it that way, I'll bear that in mind.

    Though, I'm still no further with the original question. I've added some of the terms together and recognise that the answer you provided does also produce that, but I don't see how?
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  8. #8
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    Quote Originally Posted by Leodinas View Post
    Thank you for that, I never would have thought to do it that way, I'll bear that in mind.

    Though, I'm still no further with the original question. I've added some of the terms together and recognise that the answer you provided does also produce that, but I don't see how?
    Do the arithmetic and substract the like terms. There will only be two terms remaining after you do this.
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  9. #9
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    (\frac{1}{n}-\frac{1}{n+1})+(\frac{1}{n+1}-\frac{1}{n+2})+(\frac{1}{n+2}-\frac{1}{n+3})

    I see that these cancel, well most of them, but I don't see where the 1 or the 1/n+1 would come from.

    Thanks for your continued, rapid response; I realise it must be a pain.
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  10. #10
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    Apologies, the obvious has finally dawned on me!

    Thanks again
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