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# PS with series

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Director
Joined: 07 Jun 2004
Posts: 610

Kudos [?]: 922 [0], given: 22

Location: PA

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20 Nov 2006, 12:48
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what is the quickest way to solve such problmes

and what is the answer to the below Q

1/2 + 1/2^2 + 1/2^3 ......... 1/ 2^10

thanks

rxs0005

Kudos [?]: 922 [0], given: 22

Manager
Joined: 01 Oct 2006
Posts: 241

Kudos [?]: 14 [0], given: 0

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20 Nov 2006, 13:45
I use the geometric progression formula: a(1-r^(n+1))/ (1-r)

Eg for the above series; a = Â½, r = Â½

Â½(1-1/2^11)/1/2 = 1-(1/2^11) which is close to 1, but less than 1

Kudos [?]: 14 [0], given: 0

VP
Joined: 25 Jun 2006
Posts: 1161

Kudos [?]: 190 [0], given: 0

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20 Nov 2006, 17:46
you should know this little thing:

1/2^n + 1/2^n = 1/2^(n-1).

so you just add a 1/2^10 at the back and subtract one.
then you get 1 - 1/2^10.

Kudos [?]: 190 [0], given: 0

20 Nov 2006, 17:46
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