frac{2}{2!} + frac{2+4}{3!} + frac{2+4+6}{4!} | vedclass

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(A) The $n^{th}$ term of the series is given by $T_n = \frac{2+4+6+\dots+2n}{(n+1)!}$.Using the sum of the first $n$ even numbers,$\sum_{k=1}^{n} 2k =

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$e$

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(A) The $n^{th}$ term of the series is given by $T_n = \frac{2+4+6+\dots+2n}{(n+1)!}$.Using the sum of the first $n$ even numbers,$\sum_{k=1}^{n} 2k = 2 	imes \frac{n(n+1)}{2} = n(n+1)$.Thus,$T_n = \frac{n(n+1)}{(n+1)!} = \frac{n(n+1)}{(n+1)n(n-1)!} = \frac{1}{(n-1)!}$ for $n \geq 1$.The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{(n-1)!}$.Let $m = n-1$,then $S = \sum_{m=0}^{\infty} \frac{1}{m!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \dots = e$.

$\frac{2}{2!} + \frac{2+4}{3!} + \frac{2+4+6}{4!} + \dots$ is equal to

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