GMAT PREP exponents

aleph777 wrote:Found this one in the 700+ GMAT PREP doc that's floating around. The phrasing seems a little weird to me, and I don't understand how to get the correct answer.

[(-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

OA: D

I solved for E, though, thinking that (1/2)^k is equal to 2^-k, and therefore k and -k would net zero.

Thanks!

excuse my note, but I found this problem meaningless and GMATish...
just plug in the numbers from 1 to 10 to see the result
1/2 - 1/4 +1/8 ... 1/2^10 --> interval {1/4; 1/2}

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. Do only as much math as is needed.

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 increasing a little and then decreasing 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.

That was the problem! The question I was looking at was only half a question. The whole first sentence was missing... Thanks, Mitch.