This problem includes factorial. Can anybody solve this?

enter image source here

1 Answer
Feb 11, 2018

21

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 1, 5, 23, 119 with the factorials 1, 2, 6, 24, 120

It looks like:

1 xx 1! + 2 xx 2! + ... + n xx n! = (n+1)! - 1

Can we prove it?

Let P(n) be the proposition:

1 xx 1! + 2 xx 2! + ... + n xx n! = (n+1)! - 1

We find:

1 xx 1! = 1 = 2! - 1 = (1+1)! - 1

So P(1) is true.

Suppose P(n) is true for some n.

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 P(n) then P(n+1)

So by induction P(n) is true for all n >= 1

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.