frac{2}{1!} + frac{2 + 4}{2!} + frac{2 + 4 + | vedclass

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

Vedclass · Accepted Answer

$3e$

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(C) The $n^{th}$ term of the series is given by $T_n = \frac{2 + 4 + 6 + ... + 2n}{n!}$.Using the sum of the first $n$ even numbers,$2 + 4 + ... + 2n = n(n + 1)$.So,$T_n = \frac{n(n + 1)}{n!} = \frac{n + 1}{(n - 1)!} = \frac{(n - 1) + 2}{(n - 1)!} = \frac{n - 1}{(n - 1)!} + \frac{2}{(n - 1)!} = \frac{1}{(n - 2)!} + \frac{2}{(n - 1)!}$.The sum $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{(n - 2)!} + 2 \sum_{n=1}^{\infty} \frac{1}{(n - 1)!}$.Note that $\sum_{n=1}^{\infty} \frac{1}{(n-1)!} = e$ and $\sum_{n=2}^{\infty} \frac{1}{(n-2)!} = e$.Thus,$S = e + 2e = 3e$.

$\frac{2}{1!} + \frac{2 + 4}{2!} + \frac{2 + 4 + 6}{3!} + ....\infty = $

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