Question

This problem includes factorial. Can anybody solve this?

Answer 1

`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.