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For every integer k from 1 to 10 inclusive the kth terem of

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For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 31 Jan 2012, 16:58
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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
1) greater than 2
2)between 1 & 2
3) between 0.5 and 1
4)between 0.25 and 0.5
5)less than 0.25

Could someone please provide a solution to this problem ?
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 01 Feb 2012, 22:11
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gpkk 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 first 10 terms in the sequence then T is
1) greater than 2
2)between 1 & 2
3) between 0.5 and 1
4)between 0.25 and 0.5
5)less than 0.25

Could someone please provide a solution to this problem ?


To get a hang of what the question is asking, put values for k right away. Say, k = 1, k = 2 etc
You get terms such as (1/2) when k = 1, (-1/4) when k = 2 etc

T = 1/2 - 1/4 + 1/8 - 1/16 +.... + 1/512 - 1/1024 (Sum of first 10 terms)

Of course GMAT doesn't expect us to calculate but figure out the answer using some shrewdness.

We have 10 terms. If we couple them up, two terms each, we get 5 groups:
T = (1/2 - 1/4) + (1/8 - 1/16) ...+ (1/512 - 1/1024)

Tell me, can we say that each group is positive? From a larger number, you are subtracting a smaller number in each bracket. e.g. 1/2 is larger than 1/4 so 1/2 - 1/4 = 1/4 i.e. a positive number
1/8 - 1/16 = 1/16, again a positive number.

We will get something similar to this: T = 1/4 + 1/16 +.... (all positives)
Definitely this sum, T, is greater than 1/4 i.e. 0.25

Now, let's group them in another way.

T = 1/2 + (- 1/4 + 1/8) + (- 1/16 + 1/32) ... - 1/1024
You will be able to make 4 groups since you left the first term out. The last term will also be left out.
Each bracket will give you a negative term -1/4 + 1/8 = -1/8 etc
Since the first term is 1/2 i.e. 0.5, we can say that the sum T will be less than 0.5 since all the other terms are negative.

So the sum, T, must be more than 0.25 but less than 0.5

Answer (D)
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 11 Feb 2012, 17:44
Perfect explanation!

Thanks Karishma!
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 13 Feb 2012, 06:30
VeritasPrepKarishma wrote:
gpkk 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 first 10 terms in the sequence then T is
1) greater than 2
2)between 1 & 2
3) between 0.5 and 1
4)between 0.25 and 0.5
5)less than 0.25

Could someone please provide a solution to this problem ?


To get a hang of what the question is asking, put values for k right away. Say, k = 1, k = 2 etc
You get terms such as (1/2) when k = 1, (-1/4) when k = 2 etc

T = 1/2 - 1/4 + 1/8 - 1/16 +.... + 1/512 - 1/1024 (Sum of first 10 terms)

Of course GMAT doesn't expect us to calculate but figure out the answer using some shrewdness.

We have 10 terms. If we couple them up, two terms each, we get 5 groups:
T = (1/2 - 1/4) + (1/8 - 1/16) ...+ (1/512 - 1/1024)

Tell me, can we say that each group is positive? From a larger number, you are subtracting a smaller number in each bracket. e.g. 1/2 is larger than 1/4 so 1/2 - 1/4 = 1/4 i.e. a positive number
1/8 - 1/16 = 1/16, again a positive number.

We will get something similar to this: T = 1/4 + 1/16 +.... (all positives)
Definitely this sum, T, is greater than 1/4 i.e. 0.25

Now, let's group them in another way.

T = 1/2 + (- 1/4 + 1/8) + (- 1/16 + 1/32) ... - 1/1024
You will be able to make 4 groups since you left the first term out. The last term will also be left out.
Each bracket will give you a negative term -1/4 + 1/8 = -1/8 etc
Since the first term is 1/2 i.e. 0.5, we can say that the sum T will be less than 0.5 since all the other terms are negative.

So the sum, T, must be more than 0.25 but less than 0.5

Answer (D)
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 15 Jan 2013, 06:16
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an alternative:
The series can be written as 1/2 - 1/4 + 1/8 - 1/16.............1/1024
Since the terms beyond 1/16 are too small, so I am not considering those terms.
Now on adding 1/2 - 1/4 +1/8 - 1/16, we get 5/16 which is slightly more than 4/16. Hence its value would be around `0.3.

The answer choice which includes this number is D.
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 24 Feb 2014, 17:23
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I did it the following way.

K=(1/2)-(1/4)+(1/8)-(1/16)+..... ----- 1

Multiply K by 2

2K=1-(1/2)+(1/4)-.....-(1/512) ------- 2

Adding 1 and 2

3K = 1 -(1/1024)
K= (1/3) -{1/(3*1024)}

Now 1/(3*1024) will be very small
So K= 1/3 = .3333

Ans Option D
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 30 Jun 2014, 18:37
Marcab wrote:
an alternative:
The series can be written as 1/2 - 1/4 + 1/8 - 1/16.............1/1024
Since the terms beyond 1/16 are too small, so I am not considering those terms.
Now on adding 1/2 - 1/4 +1/8 - 1/16, we get 5/16 which is slightly more than 4/16. Hence its value would be around `0.3.

The answer choice which includes this number is D.



QUESTION : WHERE IS 1/1024 coming from?
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 30 Jun 2014, 18:47
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sagnik242 wrote:
Marcab wrote:
an alternative:
The series can be written as 1/2 - 1/4 + 1/8 - 1/16.............1/1024
Since the terms beyond 1/16 are too small, so I am not considering those terms.
Now on adding 1/2 - 1/4 +1/8 - 1/16, we get 5/16 which is slightly more than 4/16. Hence its value would be around `0.3.

The answer choice which includes this number is D.



QUESTION : WHERE IS 1/1024 coming from?


When you put k = 10 (to get the last term) in the given expression (-1)^{k+1}*(1/2^k), you get -1/1024
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 12 Jul 2014, 06:54
This is GP.

The terms will be 1/2-1/4+1/8-....

Common ratio is (-1/4)/(1/2) = -1/2

So the sum of terms = 1/2 [1- (-1/2)^10]/(1-(-1/2)) = 1/2 *[1-1/1024]/3/2 = 1023/(1024*3) close to 1/3 so Option D
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 17 Jul 2014, 21:03
Can we some how solve this question using sum of a series formula- S = (n/2) × (2a + (n-1)d)??
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] New post 17 Jul 2014, 21:14
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sumitaries wrote:
Can we some how solve this question using sum of a series formula- S = (n/2) × (2a + (n-1)d)??


The formula you are talking about is used for sum of terms of an Arithmetic Progression only.

Read about it here: http://www.veritasprep.com/blog/2012/03 ... gressions/
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Re: For every integer k from 1 to 10 inclusive the kth terem of   [#permalink] 17 Jul 2014, 21:14
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