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1) For every integer K from 1 to 10 inclusive, the Kth term

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Senior Manager
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1) For every integer K from 1 to 10 inclusive, the Kth term [#permalink] New post 11 Aug 2007, 12:14
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.
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Re: Tough question [#permalink] New post 11 Aug 2007, 12:22
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))?
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 [#permalink] New post 11 Aug 2007, 12:36
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.
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 [#permalink] New post 11 Aug 2007, 13:52
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..
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Re: Tough question [#permalink] New post 11 Aug 2007, 13:56
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))?
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 [#permalink] New post 11 Aug 2007, 14:16
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!
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 [#permalink] New post 11 Aug 2007, 14:50
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.
  [#permalink] 11 Aug 2007, 14:50
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