The correct option is D 12(e^3 e) =∑_n=1^∞1+3+3^2—3^n 1n! =∑_n=1^∞3^n 12(n!) =12∑_n=1^∞((1+3^nn!) (1+1n!)) =12(e^3 e) ...

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Implicit Differentiation

11!+1+32!+1+3...

Question

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

\frac{1}{2} (e^{2} - 1)

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\frac{1}{2} (e^{3} - 1)

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\frac{1}{2} (e^{2} - e)

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\frac{1}{2} (e^{3} - e)

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Solution

The correct option is D $\frac{1}{2} (e^{3} - e)$
$= \infty \sum n = 1 \frac{1 + 3 + 3^{2} - - - 3^{n - 1}}{n!}$
$= \infty \sum n = 1 \frac{3^{n} - 1}{2 (n!)}$
$= \frac{1}{2} \infty \sum n = 1 ((1 + \frac{3^{n}}{n!}) - (1 + \frac{1}{n!}))$
$= \frac{1}{2} (e^{3} - e)$

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