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$\sum\limits _{k=1} ^{\infty}$ $\large\frac{1}{k !} \bigg(\sum \limits _{n=1}^k 2^{n-1} \bigg)=$

\[\begin {array} {1 1} (1)\;e & \quad (2)\;e^2+e \\ (3)\;e^2 & \quad (4)\;e^2-e \end {array}\]

1 Answer

(4) $e^2-e$
answered Nov 7, 2013 by pady_1
 
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