IOM Csantoshs 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
2^-k and T{1/2;1}
IOM Csantoshs 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
(-1)^(k+1) (1/2^k)= (-2)^0 * (-2^0)^k * 2^-k = 2^-ksantoshs 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
GMATGuruNY wrote:Don't compute the exact sum.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
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.