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    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF
    http://store.aqa.org.uk/qual/gce/pdf...W-MS-JAN10.PDF

    Q8B
    I don't get how on the MS they say  \dfrac{w^7 - 1}{w-1} = 0

    Can you explain ?
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    omega is a root of z^7=1 so \omega^7=1 so the above is obviously true.

    To get to that, you need to use the Geometric progression sum to N terms formula (in the formula book)
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    You need to figure out that it is a geometric series.

     \displaystyle a = 1, \ r=w, \ n=7  \\ S_n = \frac{a(r^n-1)}{r-1} = \frac{1(w^7-1)}{w-1} = \frac{w^7-1}{w-1}

    Good luck for the exam.
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    (Original post by yclicc)
    omega is a root of z^7=1 so \omega^7=1 so the above is obviously true.

    To get to that, you need to use the Geometric progression sum to N terms formula (in the formula book)

    (Original post by raheem94)
    You need to figure out that it is a geometric series.

     \displaystyle a = 1, \ r=w, \ n=7  \\ S_n = \frac{a(r^n-1)}{r-1} = \frac{1(w^7-1)}{w-1} = \frac{w^7-1}{w-1}

    Good luck for the exam.
    Thnx, I got how we got it, I just didn't see how it equaled 0, but I should have seen w^7 = 1 thus numerator = 0.

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