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    find the sum of 2n terms of the series

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    That looks like an expression rather than a series?
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    (Original post by BobBobson)
    That looks like an expression rather than a series?
    It says series im just copying what the question says
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    (Original post by Carman3)
    find the sum of 2n terms of the series

    Generalise the series in terms of n and do the sum from 1 to 2n.
    To generalise , you're going to have to multiply by (-1)^f(n) so the signs oscillate.
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    (Original post by Carman3)
    find the sum of 2n terms of the series

    Write an expression for S(2n) in terms of a summation over i=1..n of two terms. You can then expand and simplify the expression. The result falls-out quite nicely.
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    (Original post by NotNotBatman)
    Generalise the series in terms of n and do the sum from 1 to 2n.
    To generalise , you're going to have to multiply by (-1)^f(n) so the signs oscillate.
    (Original post by RogerOxon)
    Write an expression for S(2n) in terms of a summation over i=1..n of two terms. You can then expand and simplify the expression. The result falls-out quite nicely.
    Can you show me how its done?
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    (Original post by Carman3)
    Can you show me how its done?
    Here's a start:
    s_{2n}=\sum\limits_{i=1}^n [(2i-1)^2 - (2i+1)^2]
    Edit: Which is wrong .. I should be using 4i as the base and adding
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    Dont know how to go further from there


    (Original post by RogerOxon)
    Here's a start:
    s_{2n}=\sum\limits_{i=1}^n [(2i-1)^2 - (2i+1)^2]
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    (Original post by Carman3)
    Dont know how to go further from there
    Expand the squared terms and simplify.
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    (Original post by RogerOxon)
    Expand the squared terms and simplify.
    Ok i managed to get
    −8n^2 which was the right answer.. But now part b) says hence find the sum of 2n+1 terms which i dont know how to do
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    (Original post by Carman3)
    Ok i managed to get
    −8n^2 which was the right answer.. But now part b) says hence find the sum of 2n+1 terms which i dont know how to do
    You already have the sum to 2n terms, so add the additional one.
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    (Original post by RogerOxon)
    You already have the sum to 2n terms, so add the additional one.
    So what would the answer be?
 
 
 
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