Let the sum of first n terms of the series $1. (1!) + 2. (2!) + 3. (3!) + . . . . . . . . . . .$ be $(n + k)! - m$ .Find $k + m$ ?

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Solution

$T_{n} = n (n!) = [(n + 1) - 1] n! = (n + 1)! - n!$
$∴ S_{n} = (n + 1)! - 1$
$\Rightarrow k + m = 2$

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