1 + frac{4^2}{3!} + frac{4^4}{5!} + dots inft | vedclass

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(D) Let the given series be $S = 1 + \frac{4^2}{3!} + \frac{4^4}{5!} + \dots \infty$. Multiplying and dividing by $4$,we get: $S = \frac{1}{4} \left(

Vedclass · Accepted Answer

$\frac{e^4 - e^{-4}}{8}$

Vedclass · Answer

(D) Let the given series be $S = 1 + \frac{4^2}{3!} + \frac{4^4}{5!} + \dots \infty$. Multiplying and dividing by $4$,we get: $S = \frac{1}{4} \left( 4 + \frac{4^3}{3!} + \frac{4^5}{5!} + \dots \infty \right)$. We know that the expansion of $\sinh(x) = \frac{e^x - e^{-x}}{2} = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots \infty$. Substituting $x = 4$,we get: $S = \frac{1}{4} \left( \sinh(4) \right) = \frac{1}{4} \left( \frac{e^4 - e^{-4}}{2} \right) = \frac{e^4 - e^{-4}}{8}$.

$1 + \frac{4^2}{3!} + \frac{4^4}{5!} + \dots \infty = $

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