The sum of first $n$ terms of the series $1.1! + 2.2! + 3.3! + 4.4! + . . .$ is

(n + 1)! - 1

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n! - 1

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(n - 1)! - 1

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None of these

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Solution

The correct option is A $(n + 1)! - 1$
Let $S_{n} = 1.1! + 2.2! + 3.3! + 4.4! + . . . + n . n!$
$= (2 - 1) 1! + (3 - 1) 2! + (4 - 1) 3! + . . . + ((n + 1) - 1) n!$
$= (2.1! - 1!) + (3.2! - 2!) + (4.3! - 3!) + . . . + ((n + 1) n! - n!)$
$= (2! - 1!) + (3! - 2!) + (4! - 3!) + . . . + ((n + 1)! - n!)$
$= (n + 1)! - 1! = (n + 1)! - 1$

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The sum of first n terms of the series 1.1!+2.2!+3.3!+4.4!+... is

The correct option is A (n+1)!−1Let Sn=1.1!+2.2!+3.3!+4.4!+...+n.n!=(2−1)1!+(3−1)2!+(4−1)3!+...+((n+1)−1)n!=(2.1!−1!)+(3.2!−2!)+(4.3!−3!)+...+((n+1)n!−n!)=(2!−1!)+(3!−2!)+(4!−3!)+...+((n+1)!−n!)=(n+1)!−1!=(n+1)!−1

The sum of first $n$ terms of the series $1.1! + 2.2! + 3.3! + 4.4! + . . .$ is