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?
$$A.\ \ \ 2^{18}$$ $$B.\ \ \ 3\left(2\right)^{^{17}}$$ $$\left(C.\ \ \ 7(2\right)^{16}$$ $$D.3\left(2\right)^{16}$$ $$E.7\left(2\right)^{15}$$
The OA is E.
What is the formula that I should use here? Experts, can you show me how to solve this PS question? Thanks in advanced.
In the sequence 1, 2, 4, 8, 16, 32, …,
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 100
- Joined: Wed Nov 29, 2017 4:38 pm
- Thanked: 14 times
the series is 1,2,4,8,16,32,............VJesus12 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?
$$A.\ \ \ 2^{18}$$ $$B.\ \ \ 3\left(2\right)^{^{17}}$$ $$\left(C.\ \ \ 7(2\right)^{16}$$ $$D.3\left(2\right)^{16}$$ $$E.7\left(2\right)^{15}$$
The OA is E.
What is the formula that I should use here? Experts, can you show me how to solve this PS question? Thanks in advanced.
in terms of power of 2 we can rewrite it in the form
2^0,2^1,2^2,2^3,2^4,2^5,........
=> nth term = 2^(n-1)
therefore 16th, 17th ,and 18th term would be 2^15,2^16,and 2^17
and their sum = 2^15+2^16+2^17= 2^15 * (1+2+4) = 7(2^15)
hence option E is correct
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
First notice the PATTERN:VJesus12 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?
$$A.\ \ \ 2^{18}$$ $$B.\ \ \ 3\left(2\right)^{^{17}}$$ $$\left(C.\ \ \ 7(2\right)^{16}$$ $$D.3\left(2\right)^{16}$$ $$E.7\left(2\right)^{15}$$
term_1 = 1 (aka 2^0)
term_2 = 2 (aka 2^1)
term_3 = 4 (aka 2^2)
term_4 = 8 (aka 2^3)
term_5 = 16 (aka 2^4)
.
.
.
Notice that the exponent is 1 LESS THAN the term number.
So, term_16 = 2^15
term_17 = 2^16
term_18 = 2^17
We want to find the sum 2^15 + 2^16 + 2^17
We can do some factoring: 2^15 + 2^16 + 2^17 = 2^15(1 + 2^1 + 2^2)
= 2^15(1 + 2 + 4)
= 2^15(7)
= E
Cheers,
Brent
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
First notice the PATTERN:VJesus12 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?
$$A.\ \ \ 2^{18}$$ $$B.\ \ \ 3\left(2\right)^{^{17}}$$ $$\left(C.\ \ \ 7(2\right)^{16}$$ $$D.3\left(2\right)^{16}$$ $$E.7\left(2\right)^{15}$$
term_1 = 1 (aka 2^0)
term_2 = 2 (aka 2^1)
term_3 = 4 (aka 2^2)
term_4 = 8 (aka 2^3)
term_5 = 16 (aka 2^4)
.
.
.
Notice that the exponent is 1 LESS THAN the term number.
So, term_16 = 2^15
term_17 = 2^16
term_18 = 2^17
We want to find the sum 2^15 + 2^16 + 2^17
We can do some factoring: 2^15 + 2^16 + 2^17 = 2^15(1 + 2^1 + 2^2)
= 2^15(1 + 2 + 4)
= 2^15(7)
= E
Cheers,
Brent
- EconomistGMATTutor
- GMAT Instructor
- Posts: 555
- Joined: Wed Oct 04, 2017 4:18 pm
- Thanked: 180 times
- Followed by:12 members
Hello VJesus12.
Let's take a look at your question.
We can rewrite the sequence as follows: $$2^0\ \ ,\ \ \ 2^1\ ,\ \ 2^2\ ,\ 2^3\ ,\ \ 2^4\ ,\ \ \ 2^5\ ,\ \ \ 2^6\ ,\ \dots$$ This is equivalent to say that the nth term is $$a_n=2^{n-1}.$$ So, we have that $$a_{16}=2^{15},\ \ a_{17}=2^{16}\ \ \ and\ a_{18}=2^{17}.$$ Therefore $$a_{16}+a_{17}+a_{18}=2^{15}+2^{16}+2^{17}=2^{15}\left(1+2+2^2\right)=7\left(2\right)^{15}.$$ So, the correct answer is E.
I hope this explanation may help you.
Feel free to ask me again if you have a doubt.
Regards.
Let's take a look at your question.
We can rewrite the sequence as follows: $$2^0\ \ ,\ \ \ 2^1\ ,\ \ 2^2\ ,\ 2^3\ ,\ \ 2^4\ ,\ \ \ 2^5\ ,\ \ \ 2^6\ ,\ \dots$$ This is equivalent to say that the nth term is $$a_n=2^{n-1}.$$ So, we have that $$a_{16}=2^{15},\ \ a_{17}=2^{16}\ \ \ and\ a_{18}=2^{17}.$$ Therefore $$a_{16}+a_{17}+a_{18}=2^{15}+2^{16}+2^{17}=2^{15}\left(1+2+2^2\right)=7\left(2\right)^{15}.$$ So, the correct answer is E.
I hope this explanation may help you.
Feel free to ask me again if you have a doubt.
Regards.
GMAT Prep From The Economist
We offer 70+ point score improvement money back guarantee.
Our average student improves 98 points.
We offer 70+ point score improvement money back guarantee.
Our average student improves 98 points.