Question

find the sum of 1!+2!+3!+4!+...100!; find the sum of 1!+2!+3!+4!+...100!

Answer 1

That Goes with this that as 4! = 24 ( 4*3*2*1) , That's why all the numbers ahead and including 4! would be divisible by 24 thus leaving the remainder 0.

So our Question remains = 1! + 2! + 3! / 24

That is 1 + ( 2*1 ) + ( 3*2*1 ) / 24 = 9/24

So our remainder will be 9

Hope you are clear with my Answer…

Answer 2

a=1,d=1,l=100

Sn=n/2(a+l)

S100=100/2(1+100)

=50*101=5050