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Math Help - Series

  1. #1
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    Series

    Dear Sir
    I would be grateful if you can help me to find f(r) in the below questions
    thanks
    Kingsman

    Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    (not true) sorry
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  3. #3
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    Thanks .
    Incidentally there an an unwanted number 1 at the end of the series .Here the corrected series.Thanks
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  4. #4
    Super Member PaulRS's Avatar
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    What is \frac{1}{1-x^k} - \frac{1}{1-x^{k+1}} ?

    Relate it to your series.
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  5. #5
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    thanks very much for the hint but I would be grateful if you can show me how to break the function into f(k)-f(k+1).
    thanks very much.
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  6. #6
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    \displaystyle\huge\frac{x}{(1-x)\left(1-x^2\right)}+\frac{x^2}{\left(1-x^2\right)\left(1-x^3\right)}+\frac{x^3}{\left(1-x^3\right)\left(1-x^4\right)}+.....\frac{x^k}{\left(1-x^k\right)\left(1-x^{k+1}\right)}


    =\displaystyle\huge\frac{1}{1-x}\left[\frac{x}{\left(1-x^2\right)}+\frac{x^2(1-x)}{\left(1-x^2\right)\left(1-x^3\right)}+\frac{x^3(1-x)}{\left(1-x^3\right)\left(1-x^4\right)}+....\frac{x^k(1-x)}{\left(1-x^k\right)\left(1-x^{k+1}\right)}\right]


    =\displaystyle\huge\frac{1}{1-x}\left[\frac{x}{\left(1-x^2\right)}+\frac{x^2-x^3}{\left(1-x^2\right)\left(1-x^3\right)}+\frac{x^3-x^4}{\left(1-x^3\right)\left(1-x^4\right)}+...\right]

    =\displaystyle\huge\frac{1}{1-x}\left[\frac{x}{1-x^2}+\frac{x^2}{1-x^2}-\frac{x^3}{1-x^3}+\frac{x^3}{1-x^3}-\frac{x^4}{1-x^4}+\frac{x^4}{1-x^4}-.....\right]

    =\displaystyle\huge\frac{1}{1-x}\left[\frac{x(1+x)}{(1-x)(1+x)}-\frac{x^{k+1}}{1-x^{k+1}}\right]=\frac{x}{(1-x)^2}-\frac{x^{k+1}}{(1-x)\left(1-x^{k+1}\right)}
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  7. #7
    Super Member PaulRS's Avatar
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    We have \frac{1}{1-x^k}-\frac{1}{1-x^{k+1}} = \frac{x^k-x^{k+1}}{(1-x^k)\cdot (1-x^{k+1})}= (1-x)\cdot \frac{x^k}{(1-x^k)\cdot (1-x^{k+1})}.

    Hence choosing f(k)=\tfrac{1}{1-x}\cdot \frac{1}{1-x^k} works because f(k)-f(k+1)=\frac{x^k}{(1-x^k)\cdot (1-x^{k+1})}

    Then \sum_{k=1}^n\frac{x^k}{(1-x^k)\cdot (1-x^{k+1})} = \sum_{k=1}^n\left(f(k)-f(k+1)\right)=f(1)-f(n+1)=\tfrac{1}{1-x}\cdot \left(\tfrac{1}{1-x}-\tfrac{1}{1-x^{n+1}}\right)
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