Question

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

[AMU 1999; DCE 2005]

• A. 3(n!) + n - 3
• B. (n + 1)! - (n - 1)!
• C. (n + 1)! - 1
• D. 2(n!) - 2n - 1

Answer 1

The correct option is C (n + 1)! - 1

$S_n$ = 1(1!) + 2(2!) + 3(3!) + ...... + 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!.