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

Question:

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

Solution:

Let the given statement be $P(n)$, i.e.,

$\mathrm{P}(n): n(n+1)(n+5)$, which is a multiple of $3 .$

It can be noted that $P(n)$ is true for $n=1$ since $1(1+1)(1+5)=12$, which is a multiple of 3 .

Let $\mathrm{P}(k)$ be true for some positive integer $k$, i.e.,

$k(k+1)(k+5)$ is a multiple of 3

$\therefore k(k+1)(k+5)=3 m$, where $m \in \mathbf{N}$

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

$(k+1)\{(k+1)+1\}\{(k+1)+5\}$

$=(k+1)(k+2)\{(k+5)+1\}$

$=(k+1)(k+2)(k+5)+(k+1)(k+2)$

$=\{k(k+1)(k+5)+2(k+1)(k+5)\}+(k+1)(k+2)$

$=3 m+(k+1)\{2(k+5)+(k+2)\}$

$=3 m+(k+1)\{2 k+10+k+2\}$

$=3 m+(k+1)(3 k+12)$

$=3 m+3(k+1)(k+4)$

$=3\{m+(k+1)(k+4)\}=3 \times q$, where $q=\{m+(k+1)(k+4)\}$ is some natural number

Therefore, $(k+1)\{(k+1)+1\}\{(k+1)+5\}$ is a multiple of 3 .

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.