GMAT Prep1 question - Faster method to solve !

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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 1/2

Following is what I did to solve this problem:

Need to sum of first 10 terms in the sequence.
I know that odd terms in the sequence are : 1/2, 1/8, 1/32, 1/128, 1/512
I know that even terms in the sequence are: -1/4, -1/16, -1/64, -1/256, -1/1024
Subtracting each of the even terms from odd: I get: 1/4 + 1/8 + 1/32 + 1/256 + 1/1024
Adding them I get 341/1024. I realize it is slightly greather than 1/4 and less than 1/2. Hence D.

Is there a faster method to answer this question?
Source: — Problem Solving |

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by bigmonkey31 » Mon Jan 25, 2010 12:32 pm
You did exactly what I would do, except I don't think I would have solved for so many values of k... I think after solving for k=1 thru 6 or even 1 thru 4, you can understand that the value gets exponentially smaller (remember:fraction*fraction is always less than both original fractions) and trends towards 0. Hence, once you find the base of 1/4 and take into account that the odd and even values of k are canceling each other (to an extent), then you can infer D.

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by sreak1089 » Tue Jan 26, 2010 2:36 am
Perfect ! Like that !!
bigmonkey31 wrote:You did exactly what I would do, except I don't think I would have solved for so many values of k... I think after solving for k=1 thru 6 or even 1 thru 4, you can understand that the value gets exponentially smaller (remember:fraction*fraction is always less than both original fractions) and trends towards 0. Hence, once you find the base of 1/4 and take into account that the odd and even values of k are canceling each other (to an extent), then you can infer D.

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by bhumika.k.shah » Tue Jan 26, 2010 10:31 am
hi ,
i am sowree could u please elaborate !
i dint understand a word :-(

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by sanju09 » Wed Jan 27, 2010 4:17 am
bhumika.k.shah wrote:hi ,
i am sowree could u please elaborate !
i dint understand a word :-(
A geometric progression with its common ratio r given by (absolute value r) < 1, can be added to infinite terms to expect a finite answer, using the well known formula a/(1 - r), where a is the first term of the GP. Sreak got the GP ¼ + 1/8 + 1/32 + 1/256 + 1/1024, with its first term as ¼ and (absolute value r) = ½. Leave alone just five terms, even if it was to be added to infinite terms, the sum would have been (¼)/(1 - ½) = ½. Hence, ¼ < ¼ + 1/8 + 1/32 + 1/256 + 1/1024 < ½.

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