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

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Senior Manager
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for every integer k, from 1 to 10 inclusive, the kth term of [#permalink] New post 01 Aug 2007, 13:57
for every integer k, from 1 to 10 inclusive, the kth term of a sequence is given by -1^(k+1) * (1/2^k) If T is the sum of the first ten digits, T is

>2
between 1 and 2
between 1/2 and 1
between 1/4 and 1/2
less than 1/4


I know how to solve it the long way... but the GMAT doesn't want us to do that... so just wondering how others worked the problem
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 [#permalink] New post 01 Aug 2007, 14:49
I solved it the long way too - took me over 7 minutes. That won't do. I got D, btw - hope after all that work at least it's right. If anyone knows a shortcut, please post it! Otherwise this type of question will definitely be a guess for me.
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Re: GMATprep sequence [#permalink] New post 01 Aug 2007, 20:16
anonymousegmat wrote:
for every integer k, from 1 to 10 inclusive, the kth term of a sequence is given by -1^(k+1) * (1/2^k) If T is the sum of the first ten digits, T is

>2
between 1 and 2
between 1/2 and 1
between 1/4 and 1/2
less than 1/4


I know how to solve it the long way... but the GMAT doesn't want us to do that... so just wondering how others worked the problem


T = (1/2) - (1/4) + (1/8) - (1/16) + ... and so on.

The answer would be D.

There is no real need to solve this problem actually.
Lets look at the first 2 terms, 1/2 and 1/4. Their difference is 1/4. Then you add (1/8) and the difference is now between 1/2 and 1/4.
Now you subtract 1/16 and still the result is between 1/2 and 1/4.
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 [#permalink] New post 02 Aug 2007, 14:21
I get it. Because you are dealing with a sequence in which the numbers are each exponentially smaller than the preceding number, once you know it's between 1/2 and 1/4, there's no need to add the rest - the sum of the rest of the sequence won't equal the number preceding it, even if they were all positive or all negative. (I hope I'm making sense - I'm figuring this out as I go along) - so 1/8 + 1/16 + 1/32 + 1/64, ......won't ever equal 1/4.
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 [#permalink] New post 02 Aug 2007, 20:31
Term 1: -1^2 * (1/2) = 1/2
Term 2: -1^3 * (1/4) = -1/4
Term 3: -1^4 * (1/8) = 1/8

and so on. This is a geometric progression with first term 1/2, geometric ratio -1/2.

Sum of geometric progression = 1/2(1+(-1/2)^10)/(1+1/2) = 1/3(1-1/2048) = 1/3(2047/2048) = ~ 1/3

So the sum must be between 1/4 and 1/2. Ans D
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 [#permalink] New post 02 Aug 2007, 21:27
Robin in NC wrote:
I get it. Because you are dealing with a sequence in which the numbers are each exponentially smaller than the preceding number, once you know it's between 1/2 and 1/4, there's no need to add the rest - the sum of the rest of the sequence won't equal the number preceding it, even if they were all positive or all negative. (I hope I'm making sense - I'm figuring this out as I go along) - so 1/8 + 1/16 + 1/32 + 1/64, ......won't ever equal 1/4.


That is right. :-D
  [#permalink] 02 Aug 2007, 21:27
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