`f(n) = n(n^2+5) `

We have to prove that f(n) is divisible by 6.

We can use mathematical induction here.

`f(n) = n(n^2+5)`

For n = 1

`f(1) = (1)(1+5) = 6 = 6xx1 `

So when n = 1 f(n) is divisible by 6.

Let us assume that when n = p where p>1 the result is true.

`f(p) = p(p^2+5) = 6xxK` where K is a positive integer.

`f(p+1) `

`= (p+1)((p+1)^2+5)`

`= (p+1)(p^2+2p+1+5)`

`= (p+1)((p^2+5)+(2p+1))`

`= p(p^2+5)+3(p^2+p+2)`

We know that;

square of an even number is even.`(eg;2^2 = 4, 4^2 = 16)`

Square of an odd number is odd. `(eg;3^2 = 9, 5^2 = 25)`

Addition two even numbers or two odd numbers will give you a even number. `(eg;4+4 = 8, 3+5 = 8)`

Let us consider `(p^2+p+2).`

If p is even;

`p^2` is even

`p^2+p` is even

`p^2+p+2` is even.

If p is odd;

`p^2` is odd

p^2+p is even

`p^2+p+2 ` is even.

So ultimately `p^2+p+2` is even for all positive integers p.

So we can say` p^2+p+2 = 2K_1`

`f(p+1)`

`= p(p^2+5)+3(p^2+p+2)`

`= 6K+3xx2K_1`

`= 6(K+K_1)`

So f(p+1) is also divisible by 6.

*So from mathematical induction for all positive integer n; `n(n^2+5)` is divisible by 6.*

`P(n): n(n^2+1)` is divisible by 6 .

P(n): stament that says n(n^2+1) is divisible by 6

when n=1

`P(1): 1(1^2+5)=1(1+5)=1.6=6`

6 is divisible by 6. so statement P(1) is true.

Let statement is true for n=k i.e.

`P(k): k(k^2+5)` is divisible by 6 .

To prove P(n) is true when n=k+1

`P(k+1): (k+1)((k+1)^2+5)`

`=(k+1)(k^2+1+2k+5)`

`=k(k^2+5)+(k^2+1+2k+5)+k(1+2k)`

`=k(k^2+5)+(3k^2+3k+6)`

`=k(k^2+5)+3(k^2+k+2)`

`k^2+k+2` is always an even number say 2M ( M is an integer).

`k^2+k+2=2M`

Thus

`P(k+1): (k+1)((k+1)^2+5)=k(k^2+5)+3.2M`

`k(k^2+5)` is divisible by 6 since P(k) is true.

6M is divisible by 6.

Thus

`k(k^2+5)+6M` is divisible by 6

Thus P(k+1) is true when P(k) is true.

Thus P(n) is true for all n.