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In a sequence 1, 2, 4, 8, 16, 32, ... each term after the
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15 Dec 2010, 11:08

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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.

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 tems 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.

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 \(\neq{1}\).

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

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. \(2^2\)
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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the
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03 Jan 2017, 10:11

ajit257 wrote:

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)

Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the
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15 Dec 2017, 08:07

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Top Contributor

1

ajit257 wrote:

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

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Re: In a sequence 1, 2, 4, 8, 16, 32, ... each term after the &nbs
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15 Dec 2017, 08:07