1+frac{1}{3 cdot 2^2}+frac{1}{5 cdot 2^4}+fra | vedclass

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(B) Let the given series be $S = 1 + \frac{1}{3 \cdot 2^2} + \frac{1}{5 \cdot 2^4} + \frac{1}{7 \cdot 2^6} + \ldots$ We know the logarithmic expansion

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$\log _e 3$

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(B) Let the given series be $S = 1 + \frac{1}{3 \cdot 2^2} + \frac{1}{5 \cdot 2^4} + \frac{1}{7 \cdot 2^6} + \ldots$ We know the logarithmic expansion: $\log _e \left( \frac{1+x}{1-x} \right) = 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots \right)$ for $|x| Multiplying the series by $2$,we get: $2S = 2 + \frac{2}{3 \cdot 2^2} + \frac{2}{5 \cdot 2^4} + \frac{2}{7 \cdot 2^6} + \ldots = 2 + \frac{1}{3 \cdot 2} + \frac{1}{5 \cdot 2^3} + \frac{1}{7 \cdot 2^5} + \ldots$ This does not match directly. Let us rewrite the series as: $S = 2 \left[ \frac{1}{2} + \frac{(1/2)^3}{3} + \frac{(1/2)^5}{5} + \ldots \right]$ Using the expansion with $x = 1/2$: $S = \log _e \left( \frac{1 + 1/2}{1 - 1/2} \right) = \log _e \left( \frac{3/2}{1/2} \right) = \log _e 3$.

$1+\frac{1}{3 \cdot 2^2}+\frac{1}{5 \cdot 2^4}+\frac{1}{7 \cdot 2^6}+\ldots$ is equal to

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