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

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VP
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Joined: 22 Nov 2007
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For every integer k from 1 to 10, inclusive, the kth term of [#permalink] New post 03 Mar 2008, 11:00
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

greater than 2
between 1 and 2
between 1/2 and 1
between 1/4 and 1/2
less than ¼
Director
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Joined: 01 Jan 2008
Posts: 629
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Re: sequence [#permalink] New post 03 Mar 2008, 11:45
marcodonzelli 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

greater than 2
between 1 and 2
between 1/2 and 1
between 1/4 and 1/2
less than ¼


I assume a(k) = (-1)^(k)+(1/2)^k

then a(1) + ... + a(10) = -1 + 1/2 + ... + 1 + (1/2)^10 = // the first components get canceled out because there are 5 (+1) and 5 (-1), the second components remain // = (1/2) + (1/2)^2 + ... + (1/2)^10 = 1 - (1/2)^10 -> C
CEO
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Re: sequence [#permalink] New post 03 Mar 2008, 11:48
marcodonzelli 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

greater than 2
between 1 and 2
between 1/2 and 1
between 1/4 and 1/2
less than ¼


D

Essentially just do up until the 5th number. Ul see its basically 1/3. After about the 5th number it gets small enough that the next number is pretty much irrelevant.
CEO
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Re: sequence [#permalink] New post 05 Mar 2008, 10:44
whats the correct sequence formula?
what's the OA?
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re: sequence   [#permalink] 05 Mar 2008, 10:44
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