- #1
Atran
- 93
- 1
Hi, I've been trying to make another expression for,
m
[tex]\sum[/tex](k!) = f(m)
k=0
I did one expression that is,
f(m) = 2(3(4(5...((m-1)(m+1)+1)...+1)+1)+1)+1
For instance, f(4) = 2(3(4+1)+1)+1
Can you hint me on finding a finite expression for the above expression?
Thanks for help.
m
[tex]\sum[/tex](k!) = f(m)
k=0
I did one expression that is,
f(m) = 2(3(4(5...((m-1)(m+1)+1)...+1)+1)+1)+1
For instance, f(4) = 2(3(4+1)+1)+1
Can you hint me on finding a finite expression for the above expression?
Thanks for help.