Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $n (n + 1) (n + 5)$ is a multiple of 3.

Let $P(n)$ be the statement: $n(n+1)(n+5)$ is a multiple of $3$. Write $f(n)=n(n+1)(n+5)=n^3+6n^2+5n$ (expanding: $n(n+1)=n^2+n$, and $(n^2+n)(n+5)=n^3+5n^2+n^2+5n=n^3+6n^2+5n$).

Base case: For $n=1$,

a multiple of $3$. So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$, i.e. there is an integer $m$ with

We must show $f(k+1)=(k+1)^3+6(k+1)^2+5(k+1)$ is a multiple of $3$.

Compute the difference $f(k+1)-f(k)$ term by term:

Adding these: …

Divisibility induction via the difference $f(k+1)-f(k)$ after expanding $f(n)=n^3+6n^2+5n$. Showing this difference is always a multiple of $3$ lets the …

• Trying to work with the unexpanded product $n(n+1)(n+5)$ directly in the inductive step, which makes isolating a multiple of $3$ difficult; expanding to the polynomial $n^3+6n^2+5n$ first makes the difference method tractable. …