frac{e^2 + 1}{2e} = | vedclass

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(B) We know that the expansion of $e^x$ is $1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. Substituting $x = 1$,we get $e = 1 + \

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We know that the expansion of $e^x$ is $1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$. Substituting $x = 1$,we get $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \infty$. Substituting $x = -1$,we get $e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots \infty$. Adding these two series: $e + e^{-1} = 2\left( 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty \right)$. Therefore,$\frac{e + e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$. Since $\frac{e^2 + 1}{2e} = \frac{e + e^{-1}}{2}$,the correct option is $B$.

$\frac{e^2 + 1}{2e} = $

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