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

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11 Aug 2007, 12:14

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C

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45% (01:44) wrong based on 74 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

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

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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))?

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

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11 Aug 2007, 13:56

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))?

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

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11 Aug 2007, 14:16

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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