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The value of $(0.16)^{\log_{2.5}\left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \text{ to } \infty\right)}$ is equal to _______.
$\frac{1}{3} + \frac{1}{3^2} + \ldots = \frac{1/3}{1-1/3} = \frac{1}{2}$
$(0.16)^{\log_{2.5}(1/2)}$
$= (2.5)^{-2\log_{2.5}(1/2)} = \left(\frac{1}{2}\right)^{-2} = 4$
BETA
