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GMAT PREP adding exponents

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Source: — Problem Solving |
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by itsratul » Wed Jan 20, 2010 5:23 pm
Start by taking out 2 as common.... as below :

2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8 = 2( 1 + 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 )
= 2 (2+2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 ) = 2^ 2 ( 1+1+2+2^2 + 2^3 + 2^4 + 2^5 + 2^6)
.............the rest of the steps can be orally counted (seeing the trend) ans is 2^9
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by Brent@GMATPrepNow » Wed Jan 20, 2010 5:32 pm
pkw209 wrote:Fastest way to solve this one?

161) 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
Answer is A.

a. 2^9
b. 2^10
c. 2^16
d. 2^35
e. 2^37
You can also do it by pattern recognition
2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
=2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8

2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
=2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8

2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8
= 2^4 + 2^4 + 2^5 + 2^6 + 2^7 +2^8

and so on to get 2^9
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by ajith » Wed Jan 20, 2010 10:35 pm
pkw209 wrote:Fastest way to solve this one?

161) 2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 +2^8

Answer is A.

a. 2^9
b. 2^10
c. 2^16
d. 2^35
e. 2^37
It can be considered as a sum of a geometric progression (2, 2^2, 2^4... 2^8) and 2

Sum of n elements in a GP is a(r^n - 1)/(r-1)

in this case 2(2^8-)/(2-1) = 2^9-2 and adding the first 2 which we omitted we get 2^9
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