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Solution

$\log$ both sides: $\log y = \frac{1}{3}\log 9 + \frac{1}{9}\log 9 + …$ Geometric series sum = $\frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}$ $\Rightarrow \log y = \frac{1}{2}\log 9 = \log 3$ $\Rightarrow y = 3$

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Solution

We know that $e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dots$ Let $S = \dfrac{2^2}{2!} + \dfrac{2^4}{4!} + \dfrac{2^6}{6!} + \dots$ Write using even terms of $e^x$: $e^2 = 1 + 2 + \dfrac{2^2}{2!} + \dfrac{2^3}{3!} + \dfrac{2^4}{4!} + \dots$ and $e^{-2} = 1 - 2 + \dfrac{2^2}{2!} - \dfrac{2^3}{3!} + \dfrac{2^4}{4!} - \dots$ Adding both, $e^2 + e^{-2} = 2\left(1 + \dfrac{2^2}{2!} + \dfrac{2^4}{4!} + \dfrac{2^6}{6!} + \dots\right)$ So, $S = \dfrac{1}{2}\left(e^2 + e^{-2} - 2\right) = \dfrac{1}{2}\left(\dfrac{e^4 + 1 - 2e^2}{e^2}\right) = \dfrac{(e^2 - 1)^2}{2e^2}.$

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Solution

Given series = $\displaystyle 1 + \frac{1}{2! \cdot 2^2} + \frac{1}{4! \cdot 2^4} + \frac{1}{6! \cdot 2^6} + \cdots$ This is the expansion of $\dfrac{e^{1/2} + e^{-1/2}}{2} = \dfrac{e^{1/2}(1 + e^{-1})}{2} = \dfrac{e + 1}{2\sqrt{e}}$