1+frac{1+2}{2 !}+frac{1+2+2^2}{3 !}+ldots is | vedclass

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(D) The general term of the series is $T_n = \frac{1+2+2^2+\ldots+2^{n-1}}{n !}$ for $n \geq 1$.Using the sum of a geometric progression, $T_n = \frac

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$e^2-e$

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(D) The general term of the series is $T_n = \frac{1+2+2^2+\ldots+2^{n-1}}{n !}$ for $n \geq 1$.Using the sum of a geometric progression, $T_n = \frac{2^n-1}{n !}$.We can rewrite this as $T_n = \frac{2^n}{n !} - \frac{1}{n !}$.The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{2^n}{n !} - \sum_{n=1}^{\infty} \frac{1}{n !}$.Recall the expansion $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n !} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots$.Thus, $\sum_{n=1}^{\infty} \frac{2^n}{n !} = e^2 - 1$ and $\sum_{n=1}^{\infty} \frac{1}{n !} = e - 1$.Substituting these values, $S = (e^2 - 1) - (e - 1) = e^2 - e$.

$1+\frac{1+2}{2 !}+\frac{1+2+2^2}{3 !}+\ldots$ is equal to

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