In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi

Question

In a 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})

Bunuel · Accepted Answer

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.

Answer: E.

But still if you are interested:

Sum of the first  n  terms of geometric progression is given by:  sum=\frac{b*(r^{n}-1)}{r-1} , where  b  is the first term,  n  # of terms and  r  is a common ratio  
eq{1} .

Sum of infinite geometric progression with common ratio  |r|<1 , is  sum=\frac{b}{1-r} , where  b  is the first term.

Hope it helps.

KarishmaB · Answer

1st term:  1 = 2^0 
2nd term:  2^1 
3rd term:  2^2 
4th term:  2^3 
5th term:  2^4

So looking at the pattern, what will be the 16th term? It will be  2^{15} 
What about the 17th term?  2^{16} 
What about the 18th term?  2^{17}

When you add them, you get  2^{15} + 2^{16} + 2^{17}  
Now you take  2^{15}  common from the 3 terms. You are left with

2^{15}* (1 + 2 + 2^2) = 2^{15}*7

Note that  2^{16}  has 16 2s. When you take out 15 2s, you are left with a single 2. Similarly,  2^{17}  has 17 2s. When you take out 15 2s, you are left with two 2s i.e.  2^2

BrentGMATPrepNow · Answer

First notice the  PATTERN : 
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

LM · Answer

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.

Thanks very Much! This is an excellent approach.

unceldolan · Answer

16th term = 2^15 (since 2^0 = 1). Hence we need 2^15+2^16+2^17.

Now take smaller numbers: 2²+2³+2^4 = 28 = 7*(2²) (which is the first term), hence 7*(2^15) will be right. E.

sagnik2422 · Answer

I don't understand where the 2^16 and 2^17 go. and why is a16 + a17 + a18 = 2^15 + 2^16 + 2^17

Note : Sorry I can't do the subscripts for the a's

prajaktt · Answer

Pattern :  
1st term, --------------------------------------, 6th term ,... 
can be written as :
1 ,(2) ,(2*2),(2*2*2),(2*2*2*2) , (2*2*2*2*2),...

which again can be written as :
1 , 2^1, 2^2 , 2^3 , 2^4 , 2^5 ,...

Therefore , 
16th term : 2^15 ---(1)
17th term : 2^16 ---(2)
18th term : 2^17 ---(3)

Adding (1),(2) & (3)
2^15 + 2^16 + 2^17 = 2^15(1+ 2^1 + 2^2) = 2^15 ( 1+2+4) = 2^15 (7)

Ans : E

Kinshook · Answer

ajit257 
Given: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twice the previous term.
Asked: What is the sum of the 16th, 17th and 18th terms in the sequence ?
In the sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twice the previous term, therefore it is geometric progression with:

Initial term a = 1
Common ratio r = 2
nth term  t_n = a*r^{n-1} = 1*2^{n-1} = 2^{n-1}

The sum of the 16th, 17th and 18th terms in the sequence =  2^{15} + 2^{16} + 2^{17} = 2^{15} (1 + 2 + 4) = 7*2^{15}

IMO E

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