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(-1)k+1(½k).

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(-1)k+1(½k).

by Ahmed MS » Sun Jun 12, 2011 11:28 pm
(-1)k+1(½k). 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

The answer is d, but I failed understand. Could anyone help me to solve the above problem.

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by newgmattest » Mon Jun 13, 2011 12:13 am
Can you elaborate what this is defined as "(-1)k+1(½k)"?
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by Anurag@Gurome » Mon Jun 13, 2011 12:27 am
The original question is...
wrote:(-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
We don't have to determine the exact sum.
Let us compute only as much as is necessary to see the pattern.

For k = 1, -1^(1 + 1)*(1/2*1) = 1/2
For k = 2, -1^(2 + 1)*(1/2*2) = -1/4
For k = 3, -1^(3 + 1)*(1/2*3) = 1/8.
For 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 increasing up a little and then reduce a little less than it increased.

The sum of the first 2 terms is 1/4. Then the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. 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 Ahmed MS » Mon Jun 13, 2011 9:05 am
Thanks Mr. Anurag for the detail explanation. Actually I just download it from one of the link. I do not know whetehr the given problems are correct or not!

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