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Math Help - Series Problem

  1. #1
    Member jacs's Avatar
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    Series Problem

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

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

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

    and have absolutely no idea how to proceed from here


    iii. Hence, evaluate 101^2 103^2 + 105^2 107^2 + …… + 1993^2 - 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 = 1^2 + 5^2 + 9^2 + 13^2 + …… + (4n–3 )^2
    Bn = 3^2 + 7^2 + 11^2 + 15^2 + ….
    For n = 1, 2, 3,….
    i. Find the nth term of Bn.
    Tn = (4n - 1)^2

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


    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 n integers and their squares.

    Same for 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 S_n not S_{2n} . To show this compute S_1=A_1-B_1 and S_2=A_2-B_2.
    CB
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