This problem includes factorial. Can anybody solve this?
1 Answer
Explanation:
Let us look at the first few sums of the form:
#1 xx 1! + 2 xx 2! + ... + n xx n!#
We find:
#1 xx 1! = 1#
#1 xx 1! + 2 xx 2! = 5#
#1 xx 1! + 2 xx 2! + 3 xx 3! = 23#
#1 xx 1! + 2 xx 2! + 3 xx 3! + 4 xx 4! = 119#
Compare
It looks like:
#1 xx 1! + 2 xx 2! + ... + n xx n! = (n+1)! - 1#
Can we prove it?
Let
#1 xx 1! + 2 xx 2! + ... + n xx n! = (n+1)! - 1#
We find:
#1 xx 1! = 1 = 2! - 1 = (1+1)! - 1#
So
Suppose
Then:
#1 xx 1! + 2 xx 2! + ... + n xx n! + (n+1) xx (n+1)!#
#= (n+1)! - 1 + (n+1) xx (n+1)!#
#= (1+(n+1)) (n+1)! - 1#
#= ((n+1)+1)! - 1#
That is, if
So by induction
Then:
#(21! - 21)/(1 xx 1! + 2 xx 2! + ... + 19 xx 19!)= (21 xx (20 !- 1))/(20! - 1) = 21#
Remarks
Oliver Heaviside famously said "Mathematics is an experimental science, and definitions do not come first, but later on."
Note that I have not used a pat formula for the identity of factorial sum and factorial above. Instead, I tried a few values, noticed a pattern and then proved the formula.
I would recommend that you do not memorise a whole number of identities by heart. There are a few that are really useful, but in general you should be able to derive any that you need.