Question

The value of $1+1.1!+2.2!+3.3!+...+n.n!$

• A. (n+1)! + 1
• B. (n -1)! + 1
• C. (n + 1)! - 1
• D. (n + 1)!

Answer 1

Answer: (D)

n . n! = [(n+1) -1] n! = (n+1)! - n!
$\therefore$ 1+ 2! - 1! + ... + (n + 1)! - n! = (n + 1)!