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For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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31 Jan 2012, 16:58

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D

E

Difficulty:

55% (hard)

Question Stats:

62% (01:42) correct
38% (00:49) wrong based on 151 sessions

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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

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

Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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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

Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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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.
_________________

Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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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.

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
_________________

This question comes up every so often in this Forum. There are a couple of different ways of thinking about this problem, but they all require a certain degree of "math."

Without too much effort, you can deduce what the sequence is:

+1/2, -1/4, +1/8, -1/16, etc.

The "key" to solving this question quickly is to think about the terms in "sets of 2"…

1/2 - 1/4 = 1/4

Since the first term in each "set of 2" is greater than the second (negative) term, we now know that each set of 2 will be positive.

1/8 - 1/16 = 1/16

Now we know that each additional set of 2 will be significantly smaller than the prior set of 2.

1/4....1/16....1/64....etc.

Without doing all of the calculations, we know…. We have 1/4 and we'll be adding tinier and tinier fractions to it. Since there are only 10 terms in the sequence, there are only 5 sets of 2, so we won't be adding much to 1/4. Based on the answer choices, only one answer makes any sense…

Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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16 Jun 2016, 20:01

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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

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