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Question 7

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by bacali » Mon Nov 17, 2008 1:03 pm
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)


OA: E
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by dmateer25 » Mon Nov 17, 2008 1:34 pm
The terms would be 2^15, 2^16, and 2^17

Just apply the answer choices to the numbers we are already given.



A. 2^18 (4th number would be sum of first 3) does 8 = 1 + 2 + 4? No
B. 3(2^17) (3 times the 3rd number) Does 3(4) = 1 +2 + 4? No
C. 7(2^16) (7 times the 2nd number) Does 7(2) = 1 + 2 + 4? No
D. 3(2^16) (3 time 2nd number) Does 3(2) = 1 +2 +4? No
E. 7(2^15) (7 times first number) Does 7(1) = 1 +2 +4? Yes

To check: does 7(4) = 4 + 8 + 16? Yes

So answer is E.

Just my approach, but I am sure there is a better one.
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by lachlanc » Mon Nov 17, 2008 1:56 pm
My approach -

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

= 2^15 (7) = E
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by jaybrium » Mon Nov 17, 2008 2:12 pm
The first step is to realize that the terms are 2^15, 2^16, and 2^17

next I applied a technique I learned (quite recently) from the incredibly knowledgeble Cramya.

2 is a special number in that 2^x + 2^x = 2^x + 1

therefore

2^15 + 2^16 + 2^17 =

2^15 + 2^15 + 2^15 + 2^16 + 2^16 =

2^15 + 2^15 + 2^15 +2^15 + 2^15 + 2^15 + 2^15 =

7(2^15) = answer E

** thanks Cramya! keep the short cuts coming **
- Jay
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by vishubn » Mon Nov 17, 2008 5:57 pm
I am sure ! cramya ! gave this nice explanation sometime back on the repeating 2 powers sequence and playing around with them

so following the sequence 2^15, 2^16, and 2^17 will be 16th,17th and 18th

adding up we get

2^15(1+2+4)

IMO=7*2^15

Vishu
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by rmpaes » Mon Nov 17, 2008 6:18 pm
This is a geometric series.
The n-th term of a geometric sequence with initial value a1 and common ratio r is given by

an = a1r^n-1
a1 = 1
r = 2

a16 = 1 (2)^ 15
a17 = 1(2) ^16
a18 = 1(2) ^17

a16 + a17 + a18 =
2^15 + 2^16 + 2^17
Factor out 2^15
2^15(1 + 2 + 2^2)
2^15(7)
My answer would be E.
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