The sum of the infinite series 1 + 2 + frac{1 | vedclass

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(D) Let the given series be $S = 1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + \dots$We can rewrite the series by separating the

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$\frac{3e - e^{-1}}{2}$

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(D) Let the given series be $S = 1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + \dots$We can rewrite the series by separating the terms with $1$ in the numerator and terms with $2$ in the numerator:$S = (1 + \frac{1}{2!} + \frac{1}{4!} + \dots) + 2(1 + \frac{1}{3!} + \frac{1}{5!} + \dots)$Note that the standard series for $e$ and $e^{-1}$ are:$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$Thus,$\frac{e + e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \dots$ and $\frac{e - e^{-1}}{2} = 1 + \frac{1}{3!} + \frac{1}{5!} + \dots$Substituting these into the expression for $S$:$S = \frac{e + e^{-1}}{2} + 2 \left( \frac{e - e^{-1}}{2} \right)$$S = \frac{e + e^{-1} + 2e - 2e^{-1}}{2} = \frac{3e - e^{-1}}{2}$.

The sum of the infinite series $1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + \dots$ is

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