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

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Senior Manager
Joined: 17 Jul 2007
Posts: 288
Location: The 408
For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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11 Aug 2007, 13:14
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Question Stats:

56% (02:50) correct 44% (01:51) wrong based on 80 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

A. greater than 2

B. between 1 and 2

C. between 0.5 and 1

D. between 0.25 and 0.5

E. less than 0.25

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Dec 2014, 08:08, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
VP
Joined: 10 Jun 2007
Posts: 1439
Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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11 Aug 2007, 13:22
zakk wrote:
1) 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 of the sequence, then T is:

a) greater than 2
b) between 1 and 2
c) between ½ and 1
d) between ¼ and ½
e) less than ¼

I guess right, but have no idea on the concept behind this.

Can you clarify (-1)^k+1 (1/2^k)?
Is it ((-1)^(k+1)) * (1/(2^K))?
Intern
Joined: 09 Aug 2007
Posts: 33
Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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11 Aug 2007, 13:36
i interpreted (-1)^k+1 (1/2^k) as --> ((-1)^(k+1)) * (1/(2^K))

i believe the answer is D.

the first term will always be either 1 or -1, the second term will be 1/2 for k=1, then 1/2 squared or 1/4 for k=2, then 1/2 cubed, etc.

so the first ten terms will be:

1/2 - 1/4 + 1/8 - 1/16 + 1/32... - 1/1024

the values decrease with each term, so the sum of the terms is never greater than 1/2 or less than 1/4.
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12
Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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11 Aug 2007, 14:52
I get B..

basically the first term -(1)^k gets cancelled out..we have 5 even number, 5 odd numbers from 1 thru 10 inclusive...

so then basically we are adding 1+ 1/2 + 1/4 +1/8....

this sum approcaches 1..however the sum is definetly between 1 and 2..
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12
Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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11 Aug 2007, 14:56
assuming its the later case -1^(K+1)

then its basically saying that for every odd K, the term would be positive and even term would be negative, however net result would they still cancel out...in that case the sum would be 1/2+1/4+1/8...i.e C

bkk145 wrote:
zakk wrote:
1) 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 of the sequence, then T is:

a) greater than 2
b) between 1 and 2
c) between ½ and 1
d) between ¼ and ½
e) less than ¼

I guess right, but have no idea on the concept behind this.

Can you clarify (-1)^k+1 (1/2^k)?
Is it ((-1)^(k+1)) * (1/(2^K))?
Manager
Joined: 27 May 2007
Posts: 128
Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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11 Aug 2007, 15:16
I agree with emarinich - and that's a great explanation, btw!

The first nmber in the sequence is 1/2, then -1/4, 1/8, etc. It increases or decreases by an exponentially smaller number each time, so matter how far you carry it, the sum will never equal a number greater than the first number, so must be between 1/4 and 1/2. Actually this same problem was posted a week or so ago - challenging and fascinating!
Senior Manager
Joined: 17 Jul 2007
Posts: 288
Location: The 408
Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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11 Aug 2007, 15:50
emarinich wrote:
i interpreted (-1)^k+1 (1/2^k) as --> ((-1)^(k+1)) * (1/(2^K))

i believe the answer is D.

the first term will always be either 1 or -1, the second term will be 1/2 for k=1, then 1/2 squared or 1/4 for k=2, then 1/2 cubed, etc.

so the first ten terms will be:

1/2 - 1/4 + 1/8 - 1/16 + 1/32... - 1/1024

the values decrease with each term, so the sum of the terms is never greater than 1/2 or less than 1/4.

that is what the question was asking. sorry if it wasn't clear. Should have hotlinked it!

thanks for the explanation as well.
Intern
Joined: 10 Dec 2014
Posts: 40
GMAT Date: 12-30-2014
Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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19 Dec 2014, 06:21
emarinich wrote:
i interpreted (-1)^k+1 (1/2^k) as --> ((-1)^(k+1)) * (1/(2^K))

i believe the answer is D.

the first term will always be either 1 or -1, the second term will be 1/2 for k=1, then 1/2 squared or 1/4 for k=2, then 1/2 cubed, etc.

so the first ten terms will be:

1/2 - 1/4 + 1/8 - 1/16 + 1/32... - 1/1024

the values decrease with each term, so the sum of the terms is never greater than 1/2 or less than 1/4.

How can this tell that it cannot be less than 1/4??
Math Expert
Joined: 02 Sep 2009
Posts: 39701
Re: For every integer k from 1 to 10, inclusive, the kth term of a certain [#permalink]

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19 Dec 2014, 08:10
LUC999 wrote:
emarinich wrote:
i interpreted (-1)^k+1 (1/2^k) as --> ((-1)^(k+1)) * (1/(2^K))

i believe the answer is D.

the first term will always be either 1 or -1, the second term will be 1/2 for k=1, then 1/2 squared or 1/4 for k=2, then 1/2 cubed, etc.

so the first ten terms will be:

1/2 - 1/4 + 1/8 - 1/16 + 1/32... - 1/1024

the values decrease with each term, so the sum of the terms is never greater than 1/2 or less than 1/4.

How can this tell that it cannot be less than 1/4??

CHECK HERE: for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
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Re: For every integer k from 1 to 10, inclusive, the kth term of a certain   [#permalink] 19 Dec 2014, 08:10
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