1*1!+2*2!+3*3!.......................2008*2008!?

Question

HERE ! REPRESENTS FACTORIAL OF A NUMBER.
HENCE CAN ANY ONE TELL ME THIS SOLUTION?

Answer 1

1
5
23
119
719....

pattern is this : 1*1! + 2*2! + 3*3! + ... + n*n! = (n+1)!-1

Answer 2

1*1!+2*2!+3*3! = 1 + 4 + 18 = 23
It can be written in (n+1)!-1 if up to infinity.
It means the answer is infinity.

Answer 3

I think it is a very basic maths,....
1*1!+2*2!+3*3! = 1 + 4 + 18 = 23.......infinite......
In another words ...
.n represents the nos. written in (n+1)!-1..... if up to infinity.

It means the answer is infinity.

Answer 4

I'm going to agree with the first poster's assertion. My concern was whether factorial binds tighter or looser than multiplication.

Multiplication, we assume, takes precedence over addition. So the question in my mind was is the expression evaluated:

a) (1*1)! + (2*2)! + (3*3)! ...
or
b) 1*(1!) + 2*(2!) + 3*(3!)...

The cites I found, one of them below, state that "new" operations bind tighter, meaning that (b) is correct.

Answer 5

1! = 1, so 1 * 1! = 1
2! = 1*2 = 2, so 2 * 2! = 4
3! = 1*2*3 = 6, so 3 * 3! = 18

so the sum 1*1! + 2*2! + 3*3! = 23

Answer 6

Nice finding, iluxa... Let's prove it by induction.
n=1 → 1*1! = 1 = 2! - 1 check

Given sum(k=1 to n)(k*k!) = (n+1)! - 1, then
sum(k=1 to n+1)(k*k!)
= sum(k=1 to n)(k*k!) + (n+1)(n+1)!
= (n+1)! - 1 + (n+1)(n+1)!
= (n+2)(n+1)! - 1
= (n+2)! - 1
QED.

Answer 7

ha ha..stupid!!!
the time u booted ur pc..connected ur net..opened yahoo..n typed the question..u cud have easily found the answer in some book or u have easily called some of ur frend or teacher who might have answered that far more quicker.....

Answer 8

1!+2^2!+3^2+4^2!.........................

1+4!+9!+16!...............

1+24+72576+4.18*10^12....................

MATH ERROR!!!

But can you tell How many iterations???Cant be upto infinity!

Answer 9

the answer is 23, hope it helps