frac{1 + frac{2^2}{2!} + frac{2^4}{3!} + frac | vedclass

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(D) Let the numerator be $N = 1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \dots \infty$. Multiplying and dividing by $2^2$,we get $N = \fra

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(D) Let the numerator be $N = 1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \dots \infty$. Multiplying and dividing by $2^2$,we get $N = \frac{1}{2^2} \left( \frac{2^2}{1!} + \frac{2^4}{2!} + \frac{2^6}{3!} + \dots \right) = \frac{1}{4} (e^{2^2} - 1) = \frac{e^4 - 1}{4}$. Let the denominator be $D = 1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^2}{4!} + \dots \infty$. Multiplying and dividing by $2^2$,we get $D = \frac{1}{2^2} \left( 2^2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \dots \right) = \frac{1}{4} (e^2 - 1 - 2 + 2) = \frac{e^2 - 1}{4}$. Thus,the ratio is $\frac{(e^4 - 1)/4}{(e^2 - 1)/4} = \frac{e^4 - 1}{e^2 - 1} = \frac{(e^2 - 1)(e^2 + 1)}{e^2 - 1} = e^2 + 1$.

$\frac{1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \dots \infty}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^2}{4!} + \dots \infty} = $

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