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Thread: Series Problem

  1. #1
    Member jacs's Avatar
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    Series Problem

    An and Bn are two series given by:
    An = $\displaystyle 1^2 $+ $\displaystyle 5^2$ + $\displaystyle 9^2 $+ $\displaystyle 13^2$ + …… + (4n–3$\displaystyle )^2$
    Bn = $\displaystyle 3^2$ + $\displaystyle 7^2$ + $\displaystyle 11^2$ + $\displaystyle 15^2$ + ….
    For n = 1, 2, 3,….
    i. Find the nth term of Bn.
    Tn = $\displaystyle (4n - 1)^2$

    ii. If Sn = An – Bn, prove that S2n = $\displaystyle -8n^2$

    i have so far for this only
    Sn = $\displaystyle 1^2 $+ $\displaystyle 5^2$ + $\displaystyle 9^2 $+ $\displaystyle 13^2$ + …… + (4n–3$\displaystyle )^2$
    - ($\displaystyle 3^2$ + $\displaystyle 7^2$ + $\displaystyle 11^2$ + $\displaystyle 15^2$ + $\displaystyle (4n - 1)^2$)

    and have absolutely no idea how to proceed from here


    iii. Hence, evaluate $\displaystyle 101^2$ – $\displaystyle 103^2$ + $\displaystyle 105^2$ – $\displaystyle 107^2$ + …… + $\displaystyle 1993^2$ - $\displaystyle 1995^2$

    any help greatly appreciated
    thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jacs View Post
    An and Bn are two series given by:
    An = $\displaystyle 1^2 $+ $\displaystyle 5^2$ + $\displaystyle 9^2 $+ $\displaystyle 13^2$ + …… + (4n–3$\displaystyle )^2$
    Bn = $\displaystyle 3^2$ + $\displaystyle 7^2$ + $\displaystyle 11^2$ + $\displaystyle 15^2$ + ….
    For n = 1, 2, 3,….
    i. Find the nth term of Bn.
    Tn = $\displaystyle (4n - 1)^2$

    ii. If Sn = An – Bn, prove that S2n = $\displaystyle -8n^2$


    $\displaystyle A_n=\sum_{r=1}^n (4r-3)^2=\sum_{r=1}^n \left[16r^2-24r+9\right]$

    You should know the formula for the sum of the first $\displaystyle n$ integers and their squares.

    Same for $\displaystyle B_n$

    CB
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  3. #3
    Member jacs's Avatar
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    Thanks CB
    went ahead with the sigma and tried this, although i must be doing something wrong because i cannot get the required result. (pls see attached image)
    Attached Thumbnails Attached Thumbnails Series Problem-seriesproblem.png  
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by jacs View Post
    Thanks CB
    went ahead with the sigma and tried this, although i must be doing something wrong because i cannot get the required result. (pls see attached image)
    I agree, the given answer is $\displaystyle S_n$ not $\displaystyle S_{2n}$ . To show this compute $\displaystyle S_1=A_1-B_1$ and $\displaystyle S_2=A_2-B_2$.
    CB
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