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GMATPrep Q: For every integer k from 1 to 10

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GMATPrep Q: For every integer k from 1 to 10

by nandinitaneja » Sun Aug 24, 2014 7:29 am
Hello,


I saw the following question in one of my GMATPrep practice CAT's, solved it by plugging in values to the point where I realized that every odd term in the sequence was positive, and every even term was negative, and that the terms were obviously decreasing (1/2, -1/4, 1/8, -1/16 and so on). Eliminated answers A,B and C on the grounds that I was unlikely to get a number bigger than 1/2, and then guessed E as my answer. Could anyone explain a better (and fast) approach to this problem that would have lead me to the right answer?





Thank you!


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by GMATGuruNY » Sun Aug 24, 2014 7:35 am
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 first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼
Calculate until you see the pattern.
Some test-takers might find it helpful to visualize the sum on a number line.

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).
In other words, the sum will alternate between increasing a little and then decreasing a little less than it went up.

The sum of the first 2 terms is 1/4. From there, 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. Here are the first four terms, plotted on a number line:

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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 Brent@GMATPrepNow » Sun Aug 24, 2014 7:48 am
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 first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼

Plug in some values of k to see that T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
Notice that we can REWRITE this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a PATTERN emerge. We get...
T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024

Now examine the last 4 terms: 1/16 + 1/64 + 1/256 + 1/1024
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16)

Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less than 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024)
= 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Answer: D

Cheers,
Brent
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