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Thread: Series

  1. #1
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    Series

    I need help with these four questions (If possible, can it be explained in steps):

    1.
    Work out $\displaystyle \displaystyle\sum_{r=n}^{2n}\frac{r^2}{1}$

    2.
    Given that $\displaystyle f(r) \equiv 1 \frac{1}{r(r+1)}$, show that $\displaystyle f(r) - f(r+1) \equiv \frac{2}{r(r+1)(r+2)}$

    3.
    Prove that $\displaystyle \displaystyle\sum_{r=1}^{n}\frac{1}{(r+1)(r+2)}= \frac{n}{2(n+2)}$

    4.
    Find the sum of the series $\displaystyle 1^2 - 2^2 + 3^2 - 4^2 + ... - (2n)^2$
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Air View Post
    I need help with these four questions (If possible, can it be explained in steps):

    1.
    Work out $\displaystyle \displaystyle\sum_{r=n}^{2n}\frac{r^2}{1}$

    I suppose that you know that:

    $\displaystyle
    \sum_1^n r^2 = \frac{n(n+1)(2n+1)}{6}
    $

    So:

    $\displaystyle
    \sum_n^{2n} r^2 =

    $$\displaystyle \sum_{r=1}^{2n}r^2 - \sum_{r=1}^{n-1} r^2$$\displaystyle

    = \frac{2n(2n+1)(4n+1)}{6}-\frac{(n-1)(n)(2(n-1)+1)}{6}
    $

    and I leave the simplification of this to you.


    RonL
    Last edited by CaptainBlack; Sep 16th 2007 at 06:59 AM.
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  3. #3
    Math Engineering Student
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    Quote Originally Posted by Air View Post
    3.
    Prove that $\displaystyle \displaystyle\sum_{r=1}^{n}\frac{1}{(r+1)(r+2)}= \frac{n}{2(n+2)}$
    $\displaystyle \sum\limits_{r = 1}^n {\frac{1}
    {{\left( {r + 1} \right)\left( {r + 2} \right)}}} = \sum\limits_{r = 1}^n {\frac{1}
    {{r + 1}} - \frac{1}
    {{r + 2}}} = \frac{1}
    {{1 + 1}} - \frac{1}
    {{n + 2}} = \frac{n}
    {{2(n + 2)}}$
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Air View Post
    2.
    Given that $\displaystyle f(r) \equiv 1 \frac{1}{r(r+1)}$, show that $\displaystyle f(r) - f(r+1) \equiv \frac{2}{r(r+1)(r+2)}$

    $\displaystyle
    f(r) - f(r+1) = \frac{1}{r(r+1)}- \frac{1}{(r+1)(r+2)}=\frac{(r+2) - r}{r(r+1)(r+2)}=\frac{2}{r(r+1)(r+2)}
    $

    RonL
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Air View Post
    4.
    Find the sum of the series $\displaystyle 1^2 - 2^2 + 3^2 - 4^2 + ... - (2n)^2$
    Group pairs of terms:

    $\displaystyle
    \sum_{r=1}^{2n} r^2= \sum_{k=1}^{n} (2k-1)^2-(2k)^2=\sum_{k=1}^n (-4k+1)=n - 4 \sum_{k=1}^n k
    $

    Which is a sum you should be able to do.

    RonL
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