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$Value of y = (0.36)^{\log_{\frac{1}{4}} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots upto \infty)}$

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Correct option is C

13+132+…∞=131−13=13×32=12∴ y=(0.36)log14⁡12=0.3612=0.6

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Value of y=(0.36)log1413+132+133+…upto∞

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$Value of y = (0.36)^{\log_{\frac{1}{4}} (\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \dots upto \infty)}$