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Please explain , Ques 2 - 700+ Level gmat prep1

Expert replies
by dvp » Tue Oct 05, 2010 10:12 am
This is question number 2 from https://www.beatthegmat.com/198-level-70 ... 43783.html.....

Ques . (-1)^(k+1) . (1/2 ^ k). T is the sum of the first 10 k, is t
a. > 2
b. between 1 and 2
c. between ½ and 1
d. between ¼ and ½
e. < ¼

Answer : [spoiler]d (as per document)[/spoiler]

I was able to deduce the equ till ----> (-1 ) . (-1/2)^k.

how to proceed further?
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Source: — Problem Solving |

by shovan85 » Tue Oct 05, 2010 10:31 am
(-1)^(k+1) . (1/2 ^ k) plug the value of k from 1 to 10 and the equivalent sum is T

T = 1/2 - 1/4 + 1/8 - ..... + 1/512 - 1/1024 (alternate -ve from 2^-1 to 2^-10 )
=(1/2 - 1/4) + (1/8 - 1/16) + ... + (1/512 - 1/1024)
= (1/4 + 1/16) + (1/32 + 1/128) + 1/1024
= 5/16 + 5/128 + 1/1024
= 360/1024

Hence D.

Lengthy way.... any shortcut?


[/spoiler]
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by dvp » Tue Oct 05, 2010 10:44 am
Thanks Shovan,

Yes, that approach looks to take some time to solve.Maybe there is a simpler way...not sure.
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by GMATGuruNY » Tue Oct 05, 2010 11:20 am
dvp wrote:This is question number 2 from https://www.beatthegmat.com/198-level-70 ... 43783.html.....

Ques . (-1)^(k+1) . (1/2 ^ k). T is the sum of the first 10 k, is t
a. > 2
b. between 1 and 2
c. between ½ and 1
d. between ¼ and ½
e. < ¼

Answer : [spoiler]d (as per document)[/spoiler]

I was able to deduce the equ till ----> (-1 ) . (-1/2)^k.

how to proceed further?
We don't need to determine the exact sum. Compute only as much as is necessary to see the pattern.

Sum of the first two terms is 1/2 + ( -1/4) = 1/4.
Next term is 1/8.
Next term is -1/16.
Now we can see the pattern.
The sum increases by a fraction and then decreases by a fraction 1/2 the size. In other words, the sum will alternate between going up a little and then down a little less than it went up.
Since the sum of the first 2 terms is 1/4, and all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.
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by dvp » Tue Oct 05, 2010 11:13 pm
Thank you Mitch
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