Problem Solving

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...

C , right?

NO D..

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.

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?

ptgbeauregard wrote: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?

The first part does not negate itself... You get something like this:

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

--> Between 1/4 and 1/2
--> Ans is D

you're absolutely right. thanks.

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.

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

acerche wrote:how is this done in 2 minutes or less?????

it can be done, however i beleive there is a better way to solve , however i dont know it yet

(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?

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

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!

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!

loki.gmat wrote: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!

Yes Loki, this would be the preferred method. I posted a review on this topic here: https://gmatclub.com/forum/guide-to-seri ... 85969.html

When you want to find the sum of a sequence use the formula:

Sum of sequence = [First term * (1 - common ratio^Nth term)]/(1-common ratio)

Sn = a(1-r^n)/1-r, where "a" is the first term in the sequence. "a" is usually given to you. If not you can compute it with the formula for geometric series. This is posted in my review link on how to do this.

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A common mistake is that people unfamiliar with the formula will not follow PEMDAS and will subtract "r" from "1" in the numerator before taking "r to the nth power". Do not do this! You must first compute "r^n" and then subtract that from one.
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Of course on test day if you don't remember the formula or don't know it, you could always do this problem by computing it. If you know your powers of 2 then it should not take more than 2 minutes.

Thanks,
Benjiboo

Hi @GMATGuruNY,

Can you please help me solve the above question? I always find it hard to solve geometric sequences, and to be honest, I find it harder to memorize the formula! Is there a quick way to go about them without memorizing? Thank you!