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Thread: Sums of Series

  1. #1
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    Sums of Series

    Hi, well got my exam tomorrow and this is the only topic area I struggle with soo much, a past paper i'm trying here asks this qustion:

    It is given that $\displaystyle Sn=\sum(3r^2-3r+1)$

    Use the formulae for $\displaystyle \sum(r^2)$ and $\displaystyle \sum(r)$ to show that $\displaystyle Sn=n^3$.

    I know that I treat the +1 as n but I really dont know where I should take out factors or when I should expand etc. Also I get the formulas for the r squared and r so im aware of what they are.

    Each time Ive tried Ive managed to get to either $\displaystyle n^3-1$ or $\displaystyle n(n^2+1)$

    I would be really greatful if someone could explain what process I should do to get the answer.

    Thankyou !
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  2. #2
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    Quote Originally Posted by NathanBUK View Post
    Hi, well got my exam tomorrow and this is the only topic area I struggle with soo much, a past paper i'm trying here asks this qustion:

    It is given that $\displaystyle Sn=\sum(3r^2-3r+1)$

    Use the formulae for $\displaystyle \sum(r^2)$ and $\displaystyle \sum(r)$ to show that $\displaystyle Sn=n^3$.

    I know that I treat the +1 as n but I really dont know where I should take out factors or when I should expand etc. Also I get the formulas for the r squared and r so im aware of what they are.

    Each time Ive tried Ive managed to get to either $\displaystyle n^3-1$ or $\displaystyle n(n^2+1)$

    I would be really greatful if someone could explain what process I should do to get the answer.

    Thankyou !
    You probably went about it right but made some small error somewhere along the way. Here are my steps.

    $\displaystyle \sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}{6}$

    $\displaystyle \sum_{r=1}^nr=\frac{n(n+1)}{2}$

    So

    $\displaystyle \sum_{r=1}^n\left(3r^2-3r+1\right)= \sum_{r=1}^n3r^2-\sum_{r=1}^n3r+\sum_{r=1}^n1 =$

    $\displaystyle 3\sum_{r=1}^nr^2-3\sum_{r=1}^nr+\sum_{r=1}^n1 = 3\left(\frac{n(n+1)(2n+1)}{6}\right)-3\left(\frac{n(n+1)}{2}\right)+n = $

    $\displaystyle \frac{n(n+1)(2n+1)}{2}-\frac{3n(n+1)}{2}+n = \left(\frac{1}{2}\right)(n)(n+1)(-3+2n+1) + n = $

    $\displaystyle \left(\frac{1}{2}\right)(n)(n+1)(2n-2) + n = $

    $\displaystyle n(n+1)(n-1) + n = n(n^2-1) + n = n(n^2 - 1 + 1) = n(n^2) = n^3$
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  3. #3
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    Ohh thankyou ! Lets just hope one too nasty doesnt come up on my test
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