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

Author Message
Intern
Joined: 08 Nov 2008
Posts: 29
In the sequence 1, 2, 4, 8, 16, 32, , each term after the  [#permalink]

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14 Nov 2008, 09:40
15
00:00

Difficulty:

25% (medium)

Question Stats:

73% (00:46) correct 27% (00:42) wrong based on 204 sessions

### HideShow timer Statistics

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)

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-sequence-1-2-4-8-16-32-each-term-after-the-106322.html
SVP
Joined: 30 Apr 2008
Posts: 1820
Location: Oklahoma City
Schools: Hard Knocks

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14 Nov 2008, 09:55
3
E

If you look at the sum of the 1st, 2nd & 3rd, it's 1 + 2 + 4 = 7, then look at the sum of the 2nd, 3rd & 4th = 2 + 4 + 8 = 14; then 3rd, 4th & 5th = 4 + 8 + 16 = 28...there's a pattern here

so we have

7 -> the 1st through 3rd total, also known as $$7 * (2^0)$$
14 -> the 2nd through 4th, AKA $$7*(2^1)$$
28 -> the 3rd through 5th AKA $$7 * (2^2)$$

Seeing a pattern? Looking at the answers, we figure the one with 7* will be the correct answer, but which of those 2 choices will it be? let n be the nth number where we start to total. The forumula will be $$7 * 2^{n-1}$$

Since we're asked to find the 16th, 17th & 18th, and n is the nth number where we begin to start totalling, n = 16

$$7 (2^{16-1})$$ or $$7 (2^{15})$$

petercao 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)

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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VP
Joined: 05 Jul 2008
Posts: 1266

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15 Nov 2008, 11:41
Yep! Another E

1 st term is 2 ^ 0, 2nd is 2 ^1 and so on..
Intern
Joined: 28 Jul 2008
Posts: 27

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16 Nov 2008, 00:04
2
1
IMO = E

1st term: 2^0
2nd term: 2^1
.
.
.
16th term: 2^15
17th term: 2^16
18th term: 2^17

=> 2^15 + 2^16 + 2^17
= 2^15(1 + 2^1 + 2^2)
= 2^15(1 + 2 + 4)
= 7 (2^15)
Math Expert
Joined: 02 Sep 2009
Posts: 52387
Re: In the sequence 1, 2, 4, 8, 16, 32, , each term after the  [#permalink]

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21 Jan 2014, 16:58
2
4
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)

Given:
$$a_1=2^0=1$$;
$$a_2=2^1=2$$;
$$a_3=2^2=4$$;
...
$$a_n=2^{n-1}$$;

Thus $$a_{16}+a_{17}+a_{18}=2^{15}+2^{16}+2^{17}=2^{15}(1+2+4)=7*2^{15}$$.

So you don't actually need geometric series formula.

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-sequence-1-2-4-8-16-32-each-term-after-the-106322.html
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 9463
Re: In the sequence 1, 2, 4, 8, 16, 32, , each term after the  [#permalink]

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28 Jul 2018, 01:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the sequence 1, 2, 4, 8, 16, 32, , each term after the &nbs [#permalink] 28 Jul 2018, 01:14
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