1 + frac{1 + 2}{2!} + frac{1 + 2 + 3}{3!} + f | vedclass

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(D) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 3 + \dots + n}{n!} = \frac{n(n + 1)}{2 \cdot n!} = \frac{n(n + 1)}{2 \cdot n(n -

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$3e/2$

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(D) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 3 + \dots + n}{n!} = \frac{n(n + 1)}{2 \cdot n!} = \frac{n(n + 1)}{2 \cdot n(n - 1)!} = \frac{n + 1}{2(n - 1)!}$.We can rewrite the numerator as $(n - 1) + 2$,so $T_n = \frac{1}{2} \left[ \frac{n - 1}{(n - 1)!} + \frac{2}{(n - 1)!} \right] = \frac{1}{2} \left[ \frac{1}{(n - 2)!} + \frac{2}{(n - 1)!} \right]$.The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \frac{1}{2} \left[ \sum_{n=1}^{\infty} \frac{1}{(n - 2)!} + \sum_{n=1}^{\infty} \frac{2}{(n - 1)!} \right]$.Note that $\frac{1}{(n-2)!} = 0$ for $n=1$,so the first sum starts from $n=2$,which is $\sum_{k=0}^{\infty} \frac{1}{k!} = e$.The second sum is $2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} = 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 2e$.Thus,$S = \frac{1}{2} [e + 2e] = \frac{3e}{2}$.

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

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