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

santoshs 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 IOM C 2^-k and T{1/2;1}

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

santoshs 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 (-1)^(k+1) (1/2^k)= (-2)^0 * (-2^0)^k * 2^-k = 2^-k T = sum (2^-k) where k{1;10} T= 1/2 + 1/4 + 1/8 + 1/16 ... 1/1024 T=1023/1024, 1/2<T<1

GMATGuruNY wrote: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. 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.

Reminds me of geometric progressions from high school, but the way u interpreted the answer was awesome Mitch. I used the GP formula a ((r^n)-1) / (r-1) a=1/2 (1st term) r=-1/2 (factor of the GP) n=10 arrived at answer roughly equal to 1/3. Is my approach correct ?