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

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Manager
Joined: 06 Apr 2010
Posts: 117
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the  [#permalink]

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08 Sep 2010, 10:27
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35% (medium)

Question Stats:

68% (01:08) correct 32% (01:05) wrong based on 123 sessions

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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)
Senior Manager
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Re: Sequence Problem  [#permalink]

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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
we get
2^15+(2^15)*2+(2^15)*4
so answer is 2^15(1+2+4)=>7(2^15)
thanx
hope it help u!!!!
Math Expert
Joined: 02 Sep 2009
Posts: 51218
Re: Sequence Problem  [#permalink]

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08 Sep 2010, 10:44
1
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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 $$a_{18}=2^{18-1}=2^{17}$$ --> $$2^{15}+2^{16}+2^{17}=2^{15}(1+2+4)=7*2^{15}$$.

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

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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)
Manager
Joined: 06 Dec 2016
Posts: 249
Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the  [#permalink]

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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)

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

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25 Nov 2018, 21:30
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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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# In the sequence 1, 2, 4, 8, 16, 32, …, each term after the

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