Sum of the first 10 k

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Sum of the first 10 k

by missionGMAT007 » Sat Nov 20, 2010 11:42 pm
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
Last edited by missionGMAT007 on Sun Nov 21, 2010 2:38 am, edited 1 time in total.
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by Geva@EconomistGMAT » Sun Nov 21, 2010 12:42 am
Not clear - are the ks in exponents, or multiplications?

Is it (-1)*k, or (-1)^k?
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by missionGMAT007 » Sun Nov 21, 2010 2:38 am
Hello Geva,
I have edited the question

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by Geva@EconomistGMAT » Sun Nov 21, 2010 3:25 am
Are we sure about the answer? I actually came up with C.

This sort of "amorphous" questions usually contain some sort of repeating pattern that obviates the need to find all 10 first terms. Let's try out a few first terms, look for such a pattern, and remember that we don't need an exact answer - the answer choices only indicate ranges for the sum of the first 10.

K=1 (-1)^1+1 * (1/2^1) = -1 + 1/2^1 = -1/2.
K=2 (-1)^2+1 * (1/2^2) = 1 + 1/2^2 = 1+1/4 = 1.25
K=3 (-1)^3+1 * (1/2^3) = -1 + 1/2^3 = -1 + 1/8 = -0.885.
K=4 (-1)^4+1 * (1/2^4) = 1 + 1/2^4 = 1 + 1/16 = ~1.06

So what have we learned? There's a difference between even and off values of k:
when k is odd, the expression is negative, but becomes closer to -1 k grows and the 1/2^k part grows smaller.
When K is even, the expression is positive, but becomes closer to 1 with the same trend.

We want the sum of the first 10 terms. Let's look for the pattern in pairs: each pair of consecutive values of k results in a sum that is greater than zero (-1/2+1.25=0.75; -0.885+1.06 is 0.21. ), as each pair is made of a negative fraction added to a positive number greater than 1. The following three pairs will be positive fractions of negligible value, as the negative fractions grow closer to -1, and the positive numbers grow closer to 1. Thus, I believe that the sum will not move much beyond 0.75+0.21=0.96, which is still smaller than 1, but it is definitely greater than 0.75 (since the first pair is already at 0.75). This is how I reached C. Unless I am still misinterpreting the expression in some way, i believe that is the answer.
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by GMATGuruNY » Sun Nov 21, 2010 3:29 am
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
This question is from GMAT Prep. In the quote above, I've placed () around the first exponent so that it accurately reflects the original question.

As Geva noted, no need to determine the exact sum. Compute only as much as is necessary to see the pattern.

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. 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. 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 Bakhtior » Mon Nov 22, 2010 7:39 am
Geva@MasterGMAT wrote:Not clear - are the ks in exponents, or multiplications?

Is it (-1)*k, or (-1)^k?
first 10 number, 10 is even number. (-1)^k equals 0. 1/2+1/2^2+...+1/2^10= (1/2*(1-(1/2)^10))/(1/2)=1-(1/2)^10
it is between 1/2 and 1
Answer C[/img][/url][/list]