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For every integer k from 1 to 10, inclusive, the kth term of

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SVP
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Joined: 21 Jul 2006
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For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 07 Dec 2007, 04:38
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A
B
C
D
E

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For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is

a) greater than 2
b) between 1 and 2
c) between 1/2 and 1
d) between 1/4 and 1/2
e) less than 1/4



Can you guys show me how to solve this? thanks
CIO
CIO
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Joined: 09 Mar 2003
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Re: PS: Sequence [#permalink] New post 07 Dec 2007, 06:56
tarek99 wrote:
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is

a) greater than 2
b) between 1 and 2
c) between 1/2 and 1
d) between 1/4 and 1/2
e) less than 1/4



Can you guys show me how to solve this? thanks


I like this question. It takes something that's potentially hard and lets us draw a nice diagram that gets to the answer easily.

First, plot out how the sequence works. Very quickly, we will see that the sequence is made up of growing powers of 1/2, alternating positive and negative:

k = 1: 1/2
k = 2: -1/4
k = 3: 1/8
k = 4: -1/16
k = 5: 1/32
etc....

now, to add these quickly, draw a number line, with 0 in the middle, and +1 and -1 on the extremes.

The first term is 1/2, so start there. Then we subtract 1/4, which drops us down to 1/4. Then we add 1/8, so we go back up, but not quite to 1/2. Then subtract 1/16, we go down, but not exactly back to 1/4, then add 1/32, and on and on. We are essentially bouncing around a point that is in between 1/4 and 1/2.

And, lo and behold, that's it. The answer is D.
SVP
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Joined: 21 Jul 2006
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Kudos [?]: 231 [0], given: 1

 [#permalink] New post 07 Dec 2007, 10:05
beautiful explanation! the OA is D
  [#permalink] 07 Dec 2007, 10:05
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