# proof by induction that n(n² + 5) can be divided by 6? `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...

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`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.