GMAT Prep Q : Exp 3

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by Night reader » Sun Jan 02, 2011 10:31 pm
santoshs wrote:1)(-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
IOM C

2^-k and T{1/2;1}

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by santoshs » Sun Jan 02, 2011 10:43 pm
I am not able to get this problem.

(-1)^k (-1) (2)^(-k) Now how to find value of "k" and based on it find value of T

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by Night reader » Sun Jan 02, 2011 10:53 pm
santoshs wrote:I am not able to get this problem.

(-1)^k (-1) (2)^(-k) Now how to find value of "k" and based on it find value of T
(-1)^(k+1) (1/2^k)= (-2)^0 * (-2^0)^k * 2^-k = 2^-k

T = sum (2^-k) where k{1;10}
T= 1/2 + 1/4 + 1/8 + 1/16 ... 1/1024
T=1023/1024, 1/2<T<1

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by GMATGuruNY » Mon Jan 03, 2011 5:13 am
GMATGuruNY wrote:
missionGMAT007 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 1st 10 terms in the sequence then T is,

a. > 2
b. between 1 and 2
c. between ½ and 1
d. between ¼ and ½
e. < ¼

OA D
Don't compute the exact sum.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (-1/16, for example).
In other words, the sum will alternate between going up a little and then down a little less than it went up.

The sum of the first 2 terms is 1/4. Since 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 bblast » Sun Jan 16, 2011 4:24 am
Reminds me of geometric progressions from high school, but the way u interpreted the answer was awesome Mitch.

I used the GP formula
a ((r^n)-1) / (r-1)

a=1/2 (1st term)
r=-1/2 (factor of the GP)
n=10

arrived at answer roughly equal to 1/3.

Is my approach correct ?
Last edited by bblast on Sun Jan 16, 2011 4:32 am, edited 1 time in total.
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by AIM GMAT » Sun Jan 16, 2011 4:30 am
Super cool approach , this is time saving way of solving and fun too .

Thanks a lot Mitch .