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Math Help - finite series

  1. #1
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    finite series

    If S=1(n)+(2n-1)+3(n-2)+...+r(n+1-r)+...+n(1)

    and T=1(n-1)+2(n-2)+3(n-3)+...+r(n-r)+...+(n-1)(1) , where n is a positive integer , prove that

    S+T=1/6n(n+1)(2n+1)


    I can see the pattern , where :

    T=1(n-1+1)+2(n-2+1)+3(n-3+1) and so on but it doesn't lead me anywhere further .

    Thanks .
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by thereddevils View Post
    If S=1(n)+(2n-1)+3(n-2)+...+r(n+1-r)+...+n(1)

    and T=1(n-1)+2(n-2)+3(n-3)+...+r(n-r)+...+(n-1)(1) , where n is a positive integer , prove that

    S+T=1/6n(n+1)(2n+1)


    I can see the pattern , where :

    T=1(n-1+1)+2(n-2+1)+3(n-3+1) and so on but it doesn't lead me anywhere further .

    Thanks .
    S=\sum_{r=1}^n r(n+1-r)
    T=\sum_{r=1}^{n-1} r(n-r)

    S+T=\sum_{r=1}^n r(n+1-r)+\sum_{r=1}^{n-1} r(n-r)
    now make the sums begin and finish at the same values :
    S+T=n+\sum_{r=1}^{\color{red}n-1} r(n+1-r)+\sum_{r=1}^{n-1} r(n-r)

    S+T=n+\sum_{r=1}^{n-1} [r(n+1-r)+r(n-r)]
    S+T=n+\sum_{r=1}^{n-1} r(n+1-r+n-r)
    S+T=n+\sum_{r=1}^{n-1} r(2n+1-2r)
    S+T=n+(2n+1) \sum_{r=1}^{n-1} r-2 \sum_{r=1}^{n-1} r^2
    S+T=n+\frac 12 (2n+1)(n-1)n-2 \cdot \frac{(n-1)n(2n-1)}{6} \quad {\color{red}\star}
    S+T=n+\frac 16 \cdot n(n-1) \cdot [3(2n+1)-2(2n-1)]
    S+T=n+\frac 16 \cdot n(n-1)(2n+5)
    S+T=\frac n6 \cdot (6+(n-1)(2n+5))
    S+T=\frac n6 \cdot (2n^2+3n+1)
    S+T=\frac{n(2n+1)(n+1)}{6}


    {\color{red}\star} is got by these formulae :
    \sum_{j=1}^k j=\frac{k(k+1)}{2}
    and \sum_{j=1}^k j^2=\frac{k(k+1)(2k+1)}{6}
    here, k=n-1



    Of course, all of these could be solved by induction
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  3. #3
    MHF Contributor
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    Maybe a little easier

     S=\sum_{r=1}^n r(n+1-r)
     T=\sum_{r=1}^{n-1} r(n-r) =\sum_{r=1}^{n} r(n-r)

    So following Moo
     T+ S = \sum_{r=1}^n r(n+1-r) + r(n-r) = \sum_{r=1}^n r(2n+1)-2r^2  = \frac{n(n+1)(2n+1)}{2} - 2\frac{n(n+1)(2n+1)}{6} giving the answer.
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