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C2 Geometric Series

The sum of the geometric series 1+r+r2+... is k times the sum of the series 1-r+r2..., where k>0. Express r in terms of k.

By equating the sums of both series, I've got (1-rn)/(1-r) = k(1-(-r)n)/(1+r), but I'm struggling from here. Help appreciated.
Reply 1
I think you should be looking at the sum to infinity.
Reply 2
Original post by BabyMaths
I think you should be looking at the sum to infinity.


The question doesn't say anything about the value of r - would you think I could assume that -1<r<1?
Thanks
Reply 3
If |r| > 1 then the sum of the series is infinity, so presumable it means not.

Also are you sure about the second series: 1-r+r2 the 'minus' ?
(edited 11 years ago)
Reply 4
Original post by TrueGrit
If |r| > 1 then the sum of the series is infinity, so presumable it means not.

Also are you sure about the second series: 1-r+r2 the 'minus' ?


Might mean you multiply by r-r
Reply 5
Original post by Imposition
Might mean you multiply by r-r


I don't know I didn't write the question...
Reply 6
Original post by TrueGrit
I don't know I didn't write the question...


More like the question most likely means a geometric series with a ratio of -r. If the minus were a plus, there wouldn't be much of a question.
Reply 7
Original post by TrueGrit
I don't know I didn't write the question...


I meant that the common ratio is r-r
I think the OP's missing a part of the question out, sum to infinity maybe.
Reply 8
Original post by Imposition
I meant that the common ratio is r-r
I think the OP's missing a part of the question out, sum to infinity maybe.


Just re-checked the question, it's actually written "the sum of the infinite geometric series 1+r+..." but I'm not sure what it means by infinte series.
Reply 9
So, it took you 13 hours to get around to reading the question properly.

:spank:

If |r|<1 then as you add terms the sum approaches some particular value. For example S = 1+1/2+1/4+1/8+1/16......

The sums are

1
1+1/2=3/2
1+1/2+1/4=7/4
1+1/2+1/4+1/8=15/8

and it's clearly approaching 2 and you can get as close to 2 as you like.
(edited 11 years ago)
Reply 10
Original post by Julii92
The sum of the geometric series 1+r+r2+... is k times the sum of the series 1-r+r2..., where k>0. Express r in terms of k.

By equating the sums of both series, I've got (1-rn)/(1-r) = k(1-(-r)n)/(1+r), but I'm struggling from here. Help appreciated.


For infinite geometric series
1+r+r2+...=n=0rn=11r\displaystyle 1+r+r^2+... =\sum_{n=0}^{\infty} r^n=\frac{1}{1-r} for |r|<1
1r+r2r3+....=n=0(1)nrn=n=0(r)n=11(r)\displaystyle 1-r+r^2-r^3+....=\sum_{n=0}^{\infty} (-1)^n \cdot r^n=\sum_{n=0}^{\infty} (-r)^n=\frac{1}{1-(-r)}
So your equation
11r=k11+r\displaystyle \frac{1}{1-r}=k\cdot \frac{1}{1+r}
Solve for r
then from |r|<1 determine which integer can be k (maybe it's only my question)
(edited 11 years ago)

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