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2 Answers

Best answer
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19 votes

Answer is (C)

\begin{align*}
E\left( \frac{1}{1+x}\right ) &= \sum_{k = 0}^{\infty} \left( \frac{1}{1+k}\right ) * \frac{\lambda^k*e^{-\lambda}}{k!} \\
 &= \frac{1}{\lambda} * \sum_{k = 0}^{\infty}  \frac{\lambda^{k+1}*e^{-\lambda}}{(k+1)!}\\
 &= \frac{e^{-\lambda}}{\lambda} * \sum_{k = 1}^{\infty}  \frac{\lambda^{k}}{k!}\\
 &= \frac{e^{-\lambda}}{\lambda} * \left( \sum_{k = 0}^{\infty}  \frac{\lambda^{k}}{k!}-1\right)\\
 &= \frac{e^{-\lambda}}{\lambda} * \left( e^{\lambda} -1\right)\\
 &= \frac{1-e^{-\lambda}}{\lambda}

\end{align*}

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