Question

How do i prove that if n is a whole number p will be that to?

`p=n/3+n^2/2+n^3/6`

Answer 1

Please see below.

Explanation:

`p=n/3+n^2/2+n^3/6`

= `(2n+3n^2+n^3)/6`

= `(n(n^2+3n+2))/6`

= `(n(n+1)(n+2))/6`

As `n` is a whole number, `p` is a product of three consecutive
whole numbers divided by `6`.

Now when we select three consecutive numbers, at least one will be even i.e. divisible by `2`

and at least one will be divisible by `3`

Therefore `n(n+1)(n+2)` will always be divisibe by `2xx3=6`

and hence `p=n/3+n^2/2+n^3/6=(n(n+1)(n+2))/6` will be a whole number.