(2+2^2+2^3+2^4+2^5+2^6+2^7)/(2^-3+2^-4+2^-5+2^-6+2^-7+2^-8+2^-9)=?
Wow this took a long time. Shortcut?
Wow this took a long time. Shortcut?
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Source: Beat The GMAT — Problem Solving |
Both numerator and denominator are in geometric progression and you can apply the sum of the geometric progression.
Is that in scope, anways, here's my method:
2^7(2^-6+.....2^1+1)/ 2^-3((2^-6+.....2^1+1).
Faster way?
2^7(2^-6+.....2^1+1)/ 2^-3((2^-6+.....2^1+1).
Faster way?
(2+2^2+2^3+2^4+2^5+2^6+2^7)=2(1+2^1+2^2+2^3+2^4+2^5+2^6)yellowho wrote:(2+2^2+2^3+2^4+2^5+2^6+2^7)/(2^-3+2^-4+2^-5+2^-6+2^-7+2^-8+2^-9)=?
Wow this took a long time. Shortcut?
divided by
(2^-3+2^-4+2^-5+2^-6+2^-7+2^-8+2^-9)=[2^6+2^5+2^4+2^3+2^2+2^1+1]/2^9
is equal to
2^9 * 2(1+2^1+2^2+2^3+2^4+2^5+2^6) <- cancel
divided by
[2^6+2^5+2^4+2^3+2^2+2^1+1] <- cancel
is equal to
2^9*2 OR 2^10
