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In a sequence 1, 2, 4, 8, 16, 32, ... each term after the [#permalink]

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15 Dec 2010, 12:08

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E

Difficulty:

25% (medium)

Question Stats:

72% (02:02) correct
28% (01:30) wrong based on 112 sessions

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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.

In the sequence 1, 2, 4, 8, 16, 32, …, each term after the f [#permalink]

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14 Feb 2011, 17:11

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)

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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