The correct option is D e^ 3 = 3^00!+31! 3^22!+3^33! = e^ 3 #160; (e^x=1+x+x^22!+x^33!⋯ ) ...

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Logarithmic Function

2-32/2!+33/3!...

Question

$2 - \frac{3^{2}}{2!} + \frac{3^{3}}{3!} - \frac{3^{4}}{4!} + \dots . . =$

e^{3}

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e^{- 3}

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e^{- 2}

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- e^{- 3}

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Solution

The correct option is D $- e^{- 3}$
$= \frac{- 3^{0}}{0!} + \frac{3}{1!} - \frac{3^{2}}{2!} + \frac{3^{3}}{3!}$
$= - e^{- 3}$ $(e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} \dots)$

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