A function satisfies f(x-y)=\frac{f(x)}{f(y)} where f(y)\neq0. If f(1)=0.5, then what is f(2)+f(3)+f(4)+f(5)+f(6) equal to?

Explanation

The given functional equation f(x-y) = \frac{f(x)}{f(y)} represents an exponential function f(x) = a^x. Since f(1) = 0.5 = \frac{1}{2}, a = \frac{1}{2}, making f(x) = (\frac{1}{2})^x. We must evaluate a geometric progression: \sum_{x=2}^{6} (\frac{1}{2})^x = (\frac{1}{2})^2 \frac{1 - (1/2)^5}{1 - 1/2} = \frac{1}{4} \frac{31/32}{1/2} = \frac{31}{64}.

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