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

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Manager
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For every integer k from 1 to 10 inclusive the kth terem of [#permalink]

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31 Jan 2012, 17:58
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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]

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01 Feb 2012, 23:11
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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

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews VP Status: Been a long time guys... Joined: 03 Feb 2011 Posts: 1420 Location: United States (NY) Concentration: Finance, Marketing GPA: 3.75 Followers: 164 Kudos [?]: 1066 [15] , given: 62 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] ### Show Tags 15 Jan 2013, 07:16 15 This post received KUDOS Expert's post 2 This post was BOOKMARKED 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. _________________ Intern Joined: 21 Dec 2011 Posts: 14 Schools: Tepper '17 (S) Followers: 0 Kudos [?]: 10 [4] , given: 7 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] ### Show Tags 24 Feb 2014, 18:23 4 This post received KUDOS 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 Current Student Joined: 03 Feb 2013 Posts: 929 Location: India Concentration: Operations, Strategy GMAT 1: 760 Q49 V44 GPA: 3.88 WE: Engineering (Computer Software) Followers: 99 Kudos [?]: 648 [4] , given: 546 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] ### Show Tags 12 Jul 2014, 07:54 4 This post received KUDOS 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 _________________ Thanks, Kinjal Struggling with GMAT ? Experience http://www.gmatify.com/ My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961 My Linkedin Handle: https://www.linkedin.com/profile/view?id=AAIAAAbtjagB9G1MrEzTRYDRGroXuSmZO0ZhoK0 Please click on Kudos, if you think the post is helpful EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 6393 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 267 Kudos [?]: 1890 [1] , given: 161 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] ### Show Tags 03 Jun 2015, 15:14 1 This post received KUDOS Expert's post Hi vijaydoli, 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… Final Answer: [Reveal] Spoiler: D GMAT assassins aren't born, they're made, Rich _________________ # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]

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11 Feb 2012, 18:44
Perfect explanation!

Thanks Karishma!
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]

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13 Feb 2012, 07:30
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

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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]

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30 Jun 2014, 19:37
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]

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30 Jun 2014, 19:47
Expert's post
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 16 Jan 2012 Posts: 3 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] ### Show Tags 17 Jul 2014, 22:03 Can we some how solve this question using sum of a series formula- S = (n/2) × (2a + (n-1)d)?? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6565 Location: Pune, India Followers: 1788 Kudos [?]: 10753 [0], given: 210 Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink] ### Show Tags 17 Jul 2014, 22:14 Expert's post 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/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]

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04 Aug 2014, 06:45
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http://tinypic.com/r/2lvk4z5/8

Please check a simpler solution to the above problem in the above image link.

Cheers.
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For every integer k from 1 to 10 inclusive the kth terem of [#permalink]

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07 Oct 2014, 15:19
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Rohitesh,
Where did u get this 1/6?
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]

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13 Dec 2014, 21:00
taleesh wrote:
Rohitesh,
Where did u get this 1/6?

He used the summation of geometric series (base 1/4), I think.
Beautiful solution.
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Re: For every integer k from 1 to 10 inclusive the kth terem of [#permalink]

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03 Jun 2015, 13:56
I used the same calculation as above, which will probably take little more than 2 minutes. Is there a simple version to solve this problem?
Re: For every integer k from 1 to 10 inclusive the kth terem of   [#permalink] 03 Jun 2015, 13:56
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