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Urgent Help Exam tomorrow Mourning

http://store.aqa.org.uk/qual/gce/pdf/AQA-MFP2-W-QP-JAN10.PDF
http://store.aqa.org.uk/qual/gce/pdf/AQA-MFP2-W-MS-JAN10.PDF

Q8B
I don't get how on the MS they say w7−1w−1=0 \dfrac{w^7 - 1}{w-1} = 0

Can you explain ?
(edited 11 years ago)
Reply 1
Avatar for yclicc
yclicc
omega is a root of z7=1z^7=1 so ω7=1\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)
(edited 11 years ago)
Reply 2
Avatar for raheem94
raheem94
13
Original post by member910132
http://store.aqa.org.uk/qual/gce/pdf/AQA-MFP2-W-QP-JAN10.PDF
http://store.aqa.org.uk/qual/gce/pdf/AQA-MFP2-W-MS-JAN10.PDF

Q8B
I don't get how on the MS they say w7−1w−1=0 \dfrac{w^7 - 1}{w-1} = 0

Can you explain ?


You need to figure out that it is a geometric series.

a=1, r=w, n=7Sn=a(rn−1)r−1=1(w7−1)w−1=w7−1w−1 \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.
(edited 11 years ago)
Reply 3
Avatar for member910132
member910132
OP
Original post by yclicc
omega is a root of z7=1z^7=1 so ω7=1\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.

a=1, r=w, n=7Sn=a(rn−1)r−1=1(w7−1)w−1=w7−1w−1 \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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