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

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Manager
Joined: 03 Oct 2008
Posts: 61
In the sequence 1, 2, 4, 8, 16, 32, , each term after the [#permalink]

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07 Oct 2008, 08:37
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)

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Joined: 29 Aug 2007
Posts: 2452

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07 Oct 2008, 09:14
albany09 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. 2^18
B. 3(2^17)
C. 7(2^16)
D. 3(2^16)
E. 7(2^15)

D.
FIRST TERM = 1
SECOND TERM = 2^1
Third Term = 2^2

16th term = 2^15
17th term = 2^16
18th term = 2^17

sum = 2^15 + 2^16 + 2^17 = 2^15 (1+2+3) = 2^15 (6) = 3 (2^16) WOW

EDITED: sum = 2^15 + 2^16 + 2^17 = 2^15 (1+2+4) = 2^15 (7)
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07 Oct 2008, 11:06
i get E..

2^15(1+2+4)=7(2^15)
Senior Manager
Joined: 29 Mar 2008
Posts: 345

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07 Oct 2008, 11:23
(E) for me as well.
2^15+2^16+2^17 = 7(2^15)

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Re: sequence problem   [#permalink] 07 Oct 2008, 11:23
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# In the sequence 1, 2, 4, 8, 16, 32, , each term after the

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