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Math Help - Sums of Series

  1. #1
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    Sums of Series

    Hi the problem that I have is to show \sum(r^2-r)=kn(n+1)(n-1)

    where k is a rational number. (Also on the sum sign there is n on top and r=1 on bottom.

    I know what I need to do to start which is show \sum r^2-\sum r=\frac{1}{6}n(n+1)(2n+1)-\frac{1}{2}n(n+1)

    but my problem is I find it really hard to realise where the common factors are and what I should, if anything expand anywere. I would be really greatful if someone could post a step by step method on how they would do this explaing what has been extracted and what has been expanded. I know its seems alot but I would be really greatful. Thankyou!
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  2. #2
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    Quote Originally Posted by NathanBUK View Post
    Hi the problem that I have is to show \sum(r^2-r)=kn(n+1)(n-1)

    where k is a rational number. (Also on the sum sign there is n on top and r=1 on bottom.

    I know what I need to do to start which is show \sum r^2-\sum r=\frac{1}{6}n(n+1)(2n+1)-\frac{1}{2}n(n+1)

    but my problem is I find it really hard to realise where the common factors are and what I should, if anything expand anywere. I would be really greatful if someone could post a step by step method on how they would do this explaing what has been extracted and what has been expanded. I know its seems alot but I would be really greatful. Thankyou!
    hi

    first off , its obvious that n(n+1) is a common factor , and that left you with

    n(n+1)[1/6(2n+1)-1/2]

    then simplify the expression in the parentheses

    n(n+1)(1/3n-1/3)

    1/3n(n+1)(n-1)

    kn(n+1)(n-1)

    where k=1/3
    Last edited by mathaddict; April 5th 2010 at 04:37 AM.
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  3. #3
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    Quote Originally Posted by NathanBUK View Post
    Hi the problem that I have is to show \sum(r^2-r)=kn(n+1)(n-1)

    where k is a rational number. (Also on the sum sign there is n on top and r=1 on bottom.

    I know what I need to do to start which is show \sum r^2-\sum r=\frac{1}{6}n(n+1)(2n+1)-\frac{1}{2}n(n+1)

    but my problem is I find it really hard to realise where the common factors are and what I should, if anything expand anywere. I would be really greatful if someone could post a step by step method on how they would do this explaing what has been extracted and what has been expanded. I know its seems alot but I would be really greatful. Thankyou!
    n(n+1) is a common factor.
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    hi

    first off , its obvious that n(n+1) is a common factor , and that left you with

    n(n+1)[1/6(2n+1)-1/3]
    thankyou for your help but I dont quite unerstand where -1/3 came from, I must sound so dumb
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  5. #5
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    Quote Originally Posted by NathanBUK View Post
    thankyou for your help but I dont quite unerstand where -1/3 came from, I must sound so dumb
    no worries , my fault . Corrected ..
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