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 + \

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

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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 $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \infty$. 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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$1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \infty = $

In the expansion of $(e^x - 1)^2$,the coefficient of $x^4$ will be

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

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$1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \infty = $

In the expansion of $(e^x - 1)^2$,the coefficient of $x^4$ will be

$b = 1 + \frac{{}^1 C_0 + {}^1 C_1}{1!} + \frac{{}^2 C_0 + {}^2 C_1 + {}^2 C_2}{2!} + \frac{{}^3 C_0 + {}^3 C_1 + {}^3 C_2 + {}^3 C_3}{3!} + \ldots$ Let $a = 1 + \frac{{}^2 C_2}{3!} + \frac{{}^3 C_2}{4!} + \frac{{}^4 C_2}{5!} + \ldots$. Then $\frac{2b}{a^2}$ is equal to:

$\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$ is equal to

$\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$ is equal to

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$b = 1 + \frac{{}^1 C_0 + {}^1 C_1}{1!} + \frac{{}^2 C_0 + {}^2 C_1 + {}^2 C_2}{2!} + \frac{{}^3 C_0 + {}^3 C_1 + {}^3 C_2 + {}^3 C_3}{3!} + \ldots$
Let $a = 1 + \frac{{}^2 C_2}{3!} + \frac{{}^3 C_2}{4!} + \frac{{}^4 C_2}{5!} + \ldots$. Then $\frac{2b}{a^2}$ is equal to: