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    Hi, I was just wondering if anyone could give me any advice on where to start with it. the aim eventually is to make an A1 poster.
    Thanks in advance!
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    (Original post by todd.nicholls)
    Hi, I was just wondering if anyone could give me any advice on where to start with it. the aim eventually is to make an A1 poster.
    Thanks in advance!
    May be userful to rewrite the denominator. Standard formula for sum of first n squares. Then partial fractions.
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    So we got the partial fractions to equal -6/n + 6/n+1 -24/2n+1.
    We're not entirely sure where to go from here, any help would be appreciated!
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    (Original post by todd.nicholls)
    So we got the partial fractions to equal -6/n + 6/n+1 -24/2n+1.
    We're not entirely sure where to go from here, any help would be appreciated!
    Think about what happens when you add together two consecutive terms of the series (in that partial fraction form).
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    (Original post by todd.nicholls)
    So we got the partial fractions to equal -6/n + 6/n+1 -24/2n+1.
    We're not entirely sure where to go from here, any help would be appreciated!
    What happened to the (-1)^(n-1) ?

    Notice that the last part is just a multiple of the given hint if the (-1)^(n-1) was there.

    First two parts look like they would virtually all cancel in a sum; but I worry about the missing (-1) ^(n-1).
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    Sorry i'll write the equation in full.
    We get (6(-1)^n-1)/n(n+1)(2n+1) = -6/n + 6/n+1 -24/2n+1
    not sure how to get from this to the part at the bottom of the question!
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    (Original post by todd.nicholls)
    Sorry i'll write the equation in full.
    We get (6(-1)^n-1)/n(n+1)(2n+1) = -6/n + 6/n+1 -24/2n+1
    not sure how to get from this to the part at the bottom of the question!
    :holmes:

    (6(-1)^{n-1})/n(n+1)(2n+1)=6(-1)^{n-1}\left[\frac{1}{n}+\frac{1}{n+1}-\frac{1}{4(2n+1)}\right]

    Can you take it from there?
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    We have found that the first 2 terms end up cancelling leaving a remainder of 1. We're still not sure how the 3rd term is linked to the hint in any way, as we are being completely thrown by the x^2n+1.
    Thanks for the help so far!
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    (Original post by todd.nicholls)
    We're still not sure how the 3rd term is linked to the hint in any way, as we are being completely thrown by the x^2n+1.
    Well what's the obvious value to set x equal to, to get something like the series you have worked out?
 
 
 
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