Question

what is 1! + 2! + 3! + ... + n!

Answer 1

in sigma notation?
it is \[\sum_{i=1}^{n}i!\]

Answer 2

this means you add the factorial up to the last value that u r given

Answer 3

yeah thats understood.
but is there a general formula to get through this?

Answer 4

there is if you write one up for it

Answer 5

can some one please help me with this? i'm desperate :(

Answer 6

an = (an-1) + n!

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

an = (an-2) +(n-1)! + n!
= (an-3) +(n-2)! +(n-1)! + n!
= (an-r) +(n-(r-1))! +(n-(r-2))!+...+(n-2)! + (n-1)! +n!

n-r = 1 when r = (n-1)

an = (an-(n-1)) +(n-((n-1)-1))! +(n-((n-1)-2))!+...
...+(n-2)! + (n-1)! +n!
= (a1) +0! +3!+... +(n-2)! + (n-1)! +n!

i wonder where if any a missed it

Answer 7

1 (1+ 2(1 +3(1 +4(1+5+...

a1 = 1 = 1
a2 = 1 + 2 = 3
a3 = 1 +2(1+3) = 9
a4 = 1 +2(1+3(1+4))) = 33
a5 = 1 +2(1+3(1+4(1+5)))) = 153

seems to work if it can be generalized

Answer 8

an = 1 +2(1+3(1+4(...1+(n-1)(1+n)))))

Answer 9

an = 1 +2(1+3(1+4(...1+(n-1)(1+n)))))

an = 1 +2(1+3(1+4(n^2) ) ) ) )

Answer 10

Um, there isn't really any formula in the usual closed form sense for the sum of first n factorials.

Answer 11

amistre64 you approach kindles new thoughts but the final anwser you gave isn't general formula. :/

Answer 12

As i said, there isn't one.

Answer 13

il y mal.