Let alpha = frac{1}{4} + frac{1}{8} + frac{1} | vedclass

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(C) First, find the values of $\alpha$ and $\beta$ using the sum of an infinite geometric progression formula $S = \frac{a}{1-r}$.For $\alpha$: $a = 1

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$8$

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(C) First, find the values of $\alpha$ and $\beta$ using the sum of an infinite geometric progression formula $S = \frac{a}{1-r}$.For $\alpha$: $a = 1/4$, $r = 1/2$, so $\alpha = \frac{1/4}{1-1/2} = 1/2$.For $\beta$: $a = 1/3$, $r = 1/3$, so $\beta = \frac{1/3}{1-1/3} = 1/2$.Now, substitute these values into the expression: $(0.2)^{\log_{\sqrt{5}}(1/2)} + (0.04)^{\log_{5}(1/2)}$.Note that $0.2 = 5^{-1}$ and $0.04 = 5^{-2}$.For the first term: $\log_{\sqrt{5}}(1/2) = \frac{\log_5(1/2)}{\log_5(5^{1/2})} = \frac{-\log_5(2)}{1/2} = -2 \log_5(2)$.Thus, $(5^{-1})^{-2 \log_5(2)} = 5^{2 \log_5(2)} = 5^{\log_5(2^2)} = 2^2 = 4$.For the second term: $(5^{-2})^{\log_5(1/2)} = (5^{\log_5(1/2)})^{-2} = (1/2)^{-2} = 2^2 = 4$.Summing the results: $4 + 4 = 8$.

Let $\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty$ and $\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \infty$. Then the value of $(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_{5}(\beta)}$ is equal to:

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