Explanation

(c)
Here, $T_{n} = \frac{1 + 2 + 3 + ... + n}{n!} = \frac{n(n + 1)}{2(n)!}\ \left\lbrack \because\sum_{}^{}n = \frac{n(n + 1)}{2} \right\rbrack$
$= \frac{(n + 1)}{2(n - 1)!} = \frac{1}{2(n - 2)!} + \frac{1}{(n - 1)!}$
$T_{1} = 0 + \frac{1}{1}$
$T_{2} = \frac{1}{2}.\frac{1}{1} + \frac{1}{1!}$
$T_{3} = \frac{1}{2}.\frac{1}{1} + \frac{1}{2!}$
$T_{4} = \frac{1}{2}.\frac{1}{2} + \frac{1}{3!}$
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$\therefore S = \Sigma T_{n}$
$= \frac{1}{2}\left( 1 + \frac{1}{1!} + \frac{1}{2!} + ... \right) + \left( 1 + \frac{1}{1!} + \frac{1}{2!} + ... \right)$
$= \frac{1}{2}e + e = \frac{3}{2}e$