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
GMAT PREP adding exponents
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Brent@GMATPrepNow
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You can also do it by pattern recognitionpkw209 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
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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ajith
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It can be considered as a sum of a geometric progression (2, 2^2, 2^4... 2^8) and 2pkw209 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
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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