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In the sequence 1, 2, 4, 8, 16, 32, , each term after the

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Senior Manager
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Joined: 02 Dec 2007
Posts: 449

Kudos [?]: 278 [0], given: 6

In the sequence 1, 2, 4, 8, 16, 32, , each term after the [#permalink]

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New post 19 Sep 2008, 10:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?

A. 2^18
B. 3(2^17)
C. 7(2^16)
D. 3(2^16)
E. 7(2^15)

Last edited by Nihit on 19 Sep 2008, 10:48, edited 1 time in total.

Kudos [?]: 278 [0], given: 6

Director
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Joined: 12 Jul 2008
Posts: 514

Kudos [?]: 162 [0], given: 0

Schools: Wharton
Re: sequence [#permalink]

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New post 19 Sep 2008, 10:44
Nihit wrote:
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?

A. 218
B. 3(217)
C. 7(216)
D. 3(216)
E. 7(215)


Please make sure you formatting is correct after you cut and paste...

E

1st term: 2^0
2nd term: 2^1
3rd term: 2^2
4th term: 2^3
etc.

16th term + 17th term + 18th term = 2^15 + 2^16 + 2^17 = (2^15)*(1 + 2 + 2^2)
= (2^15)*(1 + 2 + 4)
= 7*(2^15)

Kudos [?]: 162 [0], given: 0

SVP
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Posts: 1513

Kudos [?]: 978 [0], given: 1

Re: sequence [#permalink]

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New post 19 Sep 2008, 18:45
Agree with the same logic as zoinnk's. I choose E as well

Kudos [?]: 978 [0], given: 1

Re: sequence   [#permalink] 19 Sep 2008, 18:45
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In the sequence 1, 2, 4, 8, 16, 32, , each term after the

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