Urgent Help Exam tomorrow Mourning

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  1. member910132's Avatar
    1 30-05-2012  18:40
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    Urgent Help Exam tomorrow Mourning
    Last edited by member910132; 30-05-2012 at 19:08.
  2. yclicc's Avatar
    2 30-05-2012  18:53
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    • Posts: 34
    Re: Urgent Help Exam tomorrow Mourning
    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)
    Last edited by yclicc; 30-05-2012 at 18:54.
  3. raheem94's Avatar
    3 30-05-2012  18:58
    • TSR Demigod
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    Re: Urgent Help Exam tomorrow Mourning
    (Original post by member910132)
    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 ?
    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.
    Last edited by raheem94; 30-05-2012 at 19:00.
  4. member910132's Avatar
    4 30-05-2012  19:07
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    Re: Urgent Help Exam tomorrow Mourning
    (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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