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

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

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New post 08 Sep 2010, 10:27
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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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Re: Sequence Problem  [#permalink]

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New post 08 Sep 2010, 10:44
the sequence comes out to be 2^0,2^1,2^2,2^3 and so on...
so 16th term will be 2^15
17th term 2^16
18th term 2^17
adding all three
we get
2^15+(2^15)*2+(2^15)*4
so answer is 2^15(1+2+4)=>7(2^15)
answer E
thanx
hope it help u!!!! :)
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Re: Sequence Problem  [#permalink]

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New post 08 Sep 2010, 10:44
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udaymathapati 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(\(2^{15}\))


\(a_1=1\), \(a_2=2=2^{2-1}\), \(a_3=4=2^{3-1}\), \(a_4=8=2^{4-1}\), \(a_5=16=2^{5-1}\), \(a_6=32=2^{6-1}\), ... --> \(a_n=2^{n-1}\);

So: \(a_{16}=2^{16-1}=2^{15}\), \(a_{17}=2^{17-1}=2^{16}\) and \(a_{18}=2^{18-1}=2^{17}\) --> \(2^{15}+2^{16}+2^{17}=2^{15}(1+2+4)=7*2^{15}\).

Answer: E.
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Sequence  [#permalink]

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New post 23 Oct 2010, 05:02
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)

What is the easiest formula for this math?
Thanks in advance...
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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the  [#permalink]

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New post 04 May 2017, 08:45
udaymathapati 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)



The series is in Geometric progression.
To find nth term, Tn = a x r ^ (n-1)
So, the sum of 16th, 17th and 18th term will be
=2^15 + 2 ^16 + 2^17
=(2^15) x (1 + 2 + 2^2)
=7 x (2^15)
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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the  [#permalink]

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New post 04 May 2017, 12:18
You could even do this approach
Look at the series carefully. You will see:

2^0, 2^1, 2^2, 2 ^3
1, 2, 3, 4

16th number is 2^15
17th number is 2^16
18th number is 2^17

2^15 + 2^16 + 2^17
2^15(1 + 2^1 + 2^2)
2^15(7)

Answer is E
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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the  [#permalink]

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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the &nbs [#permalink] 25 Nov 2018, 21:30
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