# Thread: Question on the principle of induction. Proof of divisibility

1. ## Question on the principle of induction. Proof of divisibility

Use the principle of mathematical induction to prove that $n(n^2 +5)$ is divisible by 6 for all n= $Z^+$...

2. Originally Posted by karldiesen
Use the principle of mathematical induction to prove that $n(n^2 +5)$ is divisible by 6 for all n= $Z^+$...

Lets start at the point assume $n$ and show that $n \implies n+1$

assume that $6 |n(n^2+5)$

We need to show that 6 divides $(n+1)((n+1)^2+5)$

$(n+1)((n+1)^2+5)=n^3+3n^2+3n+1+5n+5$

Now grouping the terms like this we get

$(n^3+5n) + 6 +3n(n+1)$

The first group is divisible by 6 from the induction hypothsis 6 is obviously divisible by 6.

Notice that for every integer n $n(n+1)$ is even and so it can be written in the form $2k$ for some integer k.

$3n(n+1) =3 \cdot 2k=6k$ this is also divisible by 6.