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

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Manager
Manager
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For every integer k, from 1 to 10 inclusive, the kth term of [#permalink] New post 13 Oct 2007, 13:15
For every integer k, from 1 to 10 inclusive, the kth term of the sequence is given by (-1)^k+1 (1/(2^k)).
What is the sum of the 10 terms?

1) greater than 2
2) between 1 and 2
3) between 1/2 and 1
4) between 1/4 and 1/2
5) less than 1/4

Please explain. I want to know how you guys approach such problems[/img]
VP
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 [#permalink] New post 13 Oct 2007, 13:53
Please note that (-1)^k is positive when k=2,4 or negative when k=1,3 so the total when k= from 1 to 10 is zero ! (1,3,5,7,9 cancel out 2,4,6,8,10)

so you are left with:

1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/1024 = ~ 1

the answer is (B)

:)
Manager
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 [#permalink] New post 13 Oct 2007, 14:53
Well thats not the OA i have...
CEO
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Re: GMAT Prep - Sequence [#permalink] New post 13 Oct 2007, 15:09
cruiser wrote:
For every integer k, from 1 to 10 inclusive, the kth term of the sequence is given by (-1)^k+1 (1/(2^k)).
What is the sum of the 10 terms?

1) greater than 2
2) between 1 and 2
3) between 1/2 and 1
4) between 1/4 and 1/2
5) less than 1/4

Please explain. I want to know how you guys approach such problems[/img]


I get D. Although C could be it too, but im guessin D.

-1^k+1 will be even when k is odd.

1/2^k will eventually get so small we don't need to worry about it anymore.

I tried up til k=5. or 1/2^5 ---> 1/32.

When k is 1 we get an answer of 1.5.

The list is as follows:

1.5
-1.25
1.125
-1.0625
1.03125
etc...

From this we can see that 1.5-1.25 is .25, then when added the the difference of 1.125-1.0625 we get .3125.

After this u don't really need to add anymore b/c the numbers get so small they are almost irrelevant.

Like I said i didn't add any more past this b/c 1.03125- the next number will yield a small result to add to our current .3125.

So I concluded answer is D.
Manager
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 [#permalink] New post 13 Oct 2007, 15:18
OA is D.
But i am looking for a short way of solving this...
Is there a quick way or do we have to calculate....all the terms..or some as GMATBLACKBELT did..
  [#permalink] 13 Oct 2007, 15:18
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For every integer k, from 1 to 10 inclusive, the kth term of

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