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

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In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi [#permalink]
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ajit257
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)

Could some tell me the basic formula for handling geometric series. Thanks.

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

Originally posted by BrentGMATPrepNow on 15 Dec 2017, 09:07.
Last edited by BrentGMATPrepNow on 10 Jan 2020, 06:42, edited 1 time in total.
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi [#permalink]
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.
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi [#permalink]
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.
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi [#permalink]
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
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi [#permalink]
ajit257
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)

Could some tell me the basic formula for handling geometric series. Thanks.

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)

2^15 + 2^16 + 2^17 = 2^15(1+ 2^1 + 2^2) = 2^15 ( 1+2+4) = 2^15 (7)

Ans : E
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi [#permalink]
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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
Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twi [#permalink]