frac{1}{1!} + frac{1 + 2}{2!} + frac{1 + 2 + | vedclass

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 2^2 + .... + 2^{n-1}}{n!}$.Using the sum of a geometric progression,$T_n = \frac{

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$e^2 - e$

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 2^2 + .... + 2^{n-1}}{n!}$.Using the sum of a geometric progression,$T_n = \frac{2^n - 1}{(2 - 1)n!} = \frac{2^n}{n!} - \frac{1}{n!}$.Summing the series from $n=1$ to $\infty$:$S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \left( \frac{2^n}{n!} - \frac{1}{n!} \right) = \sum_{n=1}^{\infty} \frac{2^n}{n!} - \sum_{n=1}^{\infty} \frac{1}{n!}$.We know that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \sum_{n=1}^{\infty} \frac{x^n}{n!}$,so $\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x - 1$.For $x=2$,$\sum_{n=1}^{\infty} \frac{2^n}{n!} = e^2 - 1$.For $x=1$,$\sum_{n=1}^{\infty} \frac{1^n}{n!} = e^1 - 1 = e - 1$.Therefore,$S = (e^2 - 1) - (e - 1) = e^2 - e$.

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

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