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    find the sum of the series

    1^2 -2^2 + 3^2 - 4^2+.....+(2n-1)^2 - (2n)^2

    i said that is was equal to sigma(between 1 and 2n) r^2 - sigma(between 1 and n) 2r^2

    but the the answer is

    sigma (between 1 and 2n) r^2 - 8 * sigma(between 1 and n) r^"


    I really dont understand where the *8 came from could someone explain where ive gone wrong?

    thanks
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    Think of it as

    1^2 - 2^2 + ... +(2n-1)^2 - (2n)^2

    = [1^2 + 2^2 + 3^2 + ... + (2n)^2] - 2 [2^2 + 4^2 + 6^2 + ... (2n)^2]

    = [1^2 + 2^2 + 3^2 + ... + (2n)^2] - 8 [1^2 + 2^2 + 3^2 + ... n^2]

    as

    (2k)^2 = 4 (k^2).
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    I got it simplified to something like 4n^2 - n or something... dunno if that's right, I lost my message
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    1^2 - 2^2 + 3^2 - 4^2 +... + (2n - 1)^2 - (2n)^2
    is equal to [1^2 + 2^2 + 3^2 + 4^2 +... + (2n - 1)^2 + (2n)^2 ] - 2*[ 2^2 + 4^2 +... + (2n)^2 ]
    It means that Sum from 1 to 2n of r^2 subtracts Sum from 1 to n of 2*(2r)^2
    Therefore, the equation is:
    Sigma(1 to 2n) r^2 - 8*Sigma(1 to n) r^2
 
 
 
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Updated: February 6, 2005

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