In a sequence 1, 2, 4, 8, 16, 32, ... each term after the first is twice the previous term. What is the sum of...

[AAPL]

GMAT Prep

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. \(3\cdot 2^{17}\)
C. \(7\cdot 2^{16}\)
D. \(3\cdot 2^{16}\)
E. \(7\cdot 2^{15}\)

OA E

[Brent@GMATPrepNow]

GMAT Prep

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. \(3\cdot 2^{17}\)
C. \(7\cdot 2^{16}\)
D. \(3\cdot 2^{16}\)
E. \(7\cdot 2^{15}\)

OA E

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

[Scott@TargetTestPrep]

GMAT Prep

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. \(3\cdot 2^{17}\)
C. \(7\cdot 2^{16}\)
D. \(3\cdot 2^{16}\)
E. \(7\cdot 2^{15}\)

OA E

Solution:

Instead of actually calculating the 16th, 17th and 18th terms, let’s develop a formula for a_n, the nth term of the sequence. We see that

a_1 = 2^0,
a_2 = 2^1,
a_3 = 2^2,

and so on. Notice that the exponent to which we raise 2 is 1 less than the term number.

Therefore, a_n = 2^(n - 1), and hence, a_16 = 2^15, a_17 = a^16 and a_18 = 2^17. The sum of these three terms is:

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

Answer: E