GMAT Prep ?? (Kth term)

dferm · Posted: Fri May 09, 2008 7:03 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 7:53 am Post subject:

C , right?

dferm · Posted: Fri May 09, 2008 8:24 am Post subject:

NO D..

anju · Posted: Fri May 09, 2008 11:48 am 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 12:36 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 1:38 am Post subject:

ptgbeauregard · Posted: Mon May 12, 2008 5:38 am Post subject:

you're absolutely right. thanks.

ptgbeauregard · Posted: Mon May 12, 2008 6:00 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.

acerche

how is this done in 2 minutes or less?????

yezz

rohangupta83 · Posted: Sun Oct 26, 2008 10:02 am Post subject:

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

sorry but i am still getting the above result

the 1st half is negating out if I plug into the equation

(-1)^k + (1/2^k)

Can you tell me what am i doing wrong here?

cramya · Posted: Sun Oct 26, 2008 10:29 am Post subject:

Rohan,
It should 1/2-1/4+1/8

sign reverses. I have read that this is a geometric sequence where the sum approaches a certain value once u calculate sum of the first few terms. I am sure mathematicians would knoew this but may be not us Smile

If we dint know this(I dint know either took 6-10 minutes to solve just this one in my gmat prep since it was my 3rd question) its a time consuming one for sure!

loki.gmat · Posted: Sun Oct 26, 2008 10:54 am Post subject:

here is the shortcut method.
Its a geometric progression for sure.
1st term = 1/2
2nd tern = -1/4
3rd term = 1/8
common ratio = (-1/4)/(1/2) = -1/2

sum of n terms in a geometric progression = a(1-r^n)/(1-r)
where a = 1st term r=common ration n=no of terms

Sn = (1/2)[1-(-1/2)^10]/[1-(-1/2)]
which is approximately equal to 1/3.
hence IMO C.

Thanks!