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

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VP
Joined: 22 Nov 2007
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For every integer k from 1 to 10, inclusive, the kth term of [#permalink]

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03 Mar 2008, 12:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Joined: 01 Jan 2008
Posts: 626
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Kudos [?]: 180 [0], given: 1

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03 Mar 2008, 12:45
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
Joined: 29 Mar 2007
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03 Mar 2008, 12:48
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
Joined: 21 Jan 2007
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Location: New York City
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05 Mar 2008, 11:44
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, 11:44
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