Question

The sum $1(1!)+2(2!)+3(3!)+........+n(n!)$ equals

• A. $3\left(n!\right)+n-3$
• B. $\left(n+1\right)!-(n-1)!$
• C. $(n+1)!-1$
• D. $2(n!)-2n-1$

Answer 1

The correct option is C

$(n+1)!-1$

Explanation for correct option

Given, $1(1!)+2(2!)+3(3!)+........+n(n!)$.

Simplifying,

$\begin{array}{rcl}1\left(1!\right)+2\left(2!\right)+3\left(3!\right)+...& =& 1(1!)+2(2!)+3(3!)+........+n(n!)\\ & =& (2-1)\left(1\right)!+(3-1)2!+(4-1)3!........+(n+1-1)n!\\ & =& 2\times 1!-1+3\times 2!-2!+4\times 3!-3!+.........+(n+1-1)n!\\ & =& 2!-1!+3!-2!+.....\left(n+1\right)!-n!\\ & =& (n+1)!-1\end{array}$

This kind of series where all the intermediate terms get cancelled and only the terms at the extreme remain are called as telescoping series.

Hence, option C is correct.