Vicky wrote:

Find the value of 1.1! + 2.2! + 3.3! + ......+n.n!

(1) n! +1

(2) (n+1)!

(3) (n+1)!-1

(4) (n+1)!+1

(5) n! + 3

hint: one extremely short and simple way

Had to think about this one:

You can solve this easily using substitution, but that's no fun.

Let's see if we can manipulate all of the a*a! terms

a*a! = (a+1 - 1)*a! = (a+1)a! - a! = (a+1)! - a!

Hence, 1*1! + 2*2! + .... +n*n! = (2!-1!) + (3! - 2!) + .....(n + 1)! - n!.

If you add up all of the terms, you get: -1! + (n+1)! or

(n+1)! - 1

which corresponds to answer (3).

_________________

Best,

AkamaiBrah

Former Senior Instructor, Manhattan GMAT and VeritasPrep

Vice President, Midtown NYC Investment Bank, Structured Finance IT

MFE, Haas School of Business, UC Berkeley, Class of 2005

MBA, Anderson School of Management, UCLA, Class of 1993