factorial revisited

[finsisher]

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) None of these

[cubicle]

by putting answr choices back in q

3) what is Official Answer?

[ashk29]

Answer is (n+1)!-1

1!+2*2!+3*3!+4*4!....................n*n! = 1+(3-1)2!+(4-1)3!+(5-1)4!+.....................................(n+1-1)n!
= 3*2!+4*3!+5*4!+............+(n+1)*n! - [2!+3!+4!+....n!]+1

= (n+1)!-2+1 =(n+1)!-1

[700plus2010]

ashk29

Could you please explain the rules to your last step?
I understand that (n+1)*n! = (n+1)! But I don't see how to get the "-2"

[achiever01]

sum= 1.1! + 2.2! + 3.3! . . . . n.n!
adding 1 both sides

sum + 1 = 1 + 1.1! + 2.2! + 3.3! . . . . n.n!
2! + 2.2! + 3.3! . . . . n.n!
2! (1+2) + 3.3! . . . . n.n!
3! + 3.3! . . . . n.n!
3! (1+3) . . . . n.n!
4! ... n.n!
at the end you will get (n+1)!

so sum= (n+1)! - 1

[austereaman]

lets say n=3...the sum will be 1+2*2+3*3*2*1 = 1+4+18=23

put n =3 in options....only c satisfies.....for all values of n. hence C