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.
Question 7
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
-
jaybrium
- Senior | Next Rank: 100 Posts
- Posts: 37
- Joined: Wed Sep 10, 2008 12:16 pm
- Location: Glendale, AZ
- Thanked: 3 times
- Followed by:1 members
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 **
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
-
vishubn
- Legendary Member
- Posts: 574
- Joined: Sun Jun 01, 2008 8:48 am
- Location: Bangalore
- Thanked: 28 times
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
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
KILL !! DIE !! or BEAT my FEAR !!! de@D END!!
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.
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.
700+ your target then check out my 800 gmat blog here:
https://800gmatblog.zoxic.com
https://800gmatblog.zoxic.com












