GMAT Prep ?? (Kth term)

dferm · Posted: Fri May 09, 2008 8:03 am Post subject: GMAT Prep ?? (Kth term)

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 first 10 terms in the sequence, then T is

A. greater than 2
B. between 1 and 2
C. between 1/2 and 1
D. between 1/4 and 1/2
E. less than 1/4

Please help..thanks...

bia · Posted: Fri May 09, 2008 8:53 am Post subject:

C , right?

dferm · Posted: Fri May 09, 2008 9:24 am Post subject:

NO D..

anju · Posted: Fri May 09, 2008 12:48 pm Post subject:

Hi,
I belive we will need to solve this equation for each value of K and then add the sum for calculating T. That is how the answer lies between .25 and .5 which is 1/4 and 1/2. I dunno any other faster method to solve this... Any more help will be appreciated.

Thanks.

ptgbeauregard · Posted: Sat May 10, 2008 1:36 pm Post subject:

that question is either incredibly poorly worded or the answer is wrong.

the way it reads, we are looking for the sum of k1-k10, or k1+k2+k3...+k10

the first part of the equation (-1^k) should negate itself because k1=1. k2=-1, k3=1, k4=-1, and so on so that by k10 (or the sum to any even integer), it will be 0.

so that leaves the second part of the equation, 1 (1/2^k), or 1 x (1/2^k).

obviously, you can drop the 1.

so we are really just looking at the sum or (1/2)^1+(1/2)^2+...(1/2)^10

so....

(1/2)+(1/4)+(1/8 )+(1/16)+(1/32)+(1/64)+(1/128)+(1/256)+(1/512)+(1/1024)

the number starts at 1/2 and gets arbitrarily close to 1...how is it not C?

Magellan · Posted: Mon May 12, 2008 2:38 am Post subject:

ptgbeauregard · Posted: Mon May 12, 2008 6:38 am Post subject:

you're absolutely right. thanks.

ptgbeauregard · Posted: Mon May 12, 2008 7:00 am Post subject:

fyi i think my confusion came from how the question read when copied and pasted into the post. if you read the way i worked through it, i thought it was -1^k + 1(1/2^k). i actually saw this on a practice test and it read much more natiurally with raised type for the exponent, -1^(k+1)x(1/2^k).

hence the incorrect comment that the first part would negate itself.