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    Hi guys,It's a question from CIE A level further maths, I can do the first part of the question and also can find S3, but I am not sure how to calculate S-2. The examiner report says that S3 = 3 and S-2 = 1 Any help would be greatly appreciated. Thanks!
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    (Original post by i cantthinkofany)
    Hi guys,It's a question from CIE A level further maths, I can do the first part of the question and also can find S3, but I am not sure how to calculate S-2. The examiner report says that S3 = 3 and S-2 = 1 Any help would be greatly appreciated. Thanks!
    Use the deduced equation to find what the sums and products of its roots are (remember; they're not simply \alpha,\beta,\gamma anymore)

    S_{-2}=(2\alpha+1)^{-2}+(2\beta+1)^{-2}+(2\gamma+1)^{-2}

    =\frac{1}{(2\alpha+1)^{2}}+\frac  {1}{(2\beta+1)^{2}}+\frac{1}{(2 \gamma+1)^{2}}

    =\frac{\sum(2\alpha+1)^{2}(2\beta  +1)^{2}}{(2\alpha+1)^{2}(2\beta+  1)^{2}(2\gamma+1)^{2}}

    ...and you can work from there.

    Further hints:
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    Remember that:

    \sum(\alpha\beta)^2=(\sum{\alpha  \beta})^2-2\alpha\beta\gamma\sum{\alpha}

    and

    \alpha\mapsto\ 2\alpha+1
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    If you are trying to find  S_n then you could make the substitution say  z=y^n , then to find the value of  S_n you would just be looking at the coefficient of the second term of the polynomial.
    So to find  S_2 if you let  z=y^2 and then rearrange and form a polynomial equation, and look at the second term coefficient, that will be the sum  S_2 , so it will be  -b/a where's is coefficient of first term and b is coefficient of second term.
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    Hey guys, thanks for the reply. Both of the methods seem like valid ways to do it!
 
 
 
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