The sum of frac{2}{1!} + frac{6}{2!} + frac{1 | vedclass

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(D) Let the $n^{th}$ term of the series be $T_n = \frac{n(n+1)}{n!}$.We can simplify $T_n$ as follows:$T_n = \frac{n(n+1)}{n(n-1)!} = \frac{n+1}{(n-1)

Vedclass · Accepted Answer

$3e$

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(D) Let the $n^{th}$ term of the series be $T_n = \frac{n(n+1)}{n!}$.We can simplify $T_n$ as follows:$T_n = \frac{n(n+1)}{n(n-1)!} = \frac{n+1}{(n-1)!} = \frac{(n-1)+2}{(n-1)!} = \frac{n-1}{(n-1)!} + \frac{2}{(n-1)!} = \frac{1}{(n-2)!} + \frac{2}{(n-1)!}$.The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{(n-2)!} + \sum_{n=1}^{\infty} \frac{2}{(n-1)!}$.Note that $\frac{1}{(n-2)!} = 0$ for $n=1$,so the first sum starts from $n=2$.$S = \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!}$.Using the expansion $e = \sum_{k=0}^{\infty} \frac{1}{k!}$,we get $S = e + 2e = 3e$.

The sum of $\frac{2}{1!} + \frac{6}{2!} + \frac{12}{3!} + \frac{20}{4!} + \dots$ is

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