• # question_answer $\frac{1}{1\,!}+\frac{4}{2\,!}+\frac{7}{3\,!}+\frac{10}{4\,!}+.....\infty =$ A) $e+4$ B) $2+e$ C) $3+e$ D) $e$

Solution :

$S=\frac{1}{1\ !}+\frac{4}{2\ !}+\frac{7}{3\ !}+\frac{10}{4\ !}+......+\frac{3n-2}{n\ !}+.......\infty$ Here ${{T}_{n}}=\frac{3}{(n-1)\ !}-\frac{2}{n\ !}$ $\Rightarrow S=\sum\limits_{n=1}^{\infty }{{{T}_{n}}=3\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)\ !}-2\sum\limits_{n=1}^{\infty }{\frac{1}{n\ !}}}}$        $=3e-2(e-1)=e+2$.

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