Question

show that the product of three consecutive natural number is divisible by 6​

Answer 1

The concept given below always holds true that,

$\displaystyle 2 \mid n(n+1)$

We have to prove that,

$6 \mid n(n+1)(n+2)$

where  $n \in \mathbb{N}$

Let n = 1.

$\begin{aligned}&n(n+1)(n+2)\\ \\ \Longrightarrow\ \ &1 \times 2 \times 3\\ \\ \Longrightarrow\ \ &6\end{aligned}\\ \\ \\ \\ \underline{\underline{6 \mid 6}}$

Let n = 2.

$\begin{aligned}&n(n+1)(n+2)\\ \\ \Longrightarrow\ \ &2 \times 3 \times 4\\ \\ \Longrightarrow\ \ &24\end{aligned}\\ \\ \\ \\ \underline{\underline{6 \mid 24}}$

Let n = 3.

$\begin{aligned}&n(n+1)(n+2)\\ \\ \Longrightarrow\ \ &3 \times 4 \times 5\\ \\ \Longrightarrow\ \ &60\end{aligned}\\ \\ \\ \\ \underline{\underline{6 \mid 60}}$

Let n = k.

Assume that  $6 \mid k(k+1)(k+2)$.

Taking $k(k+1)(k+2)=6m$ for any integer $m$.

Let n = k + 1.

$\begin{aligned}&n(n+1)(n+2)\\ \\ \Longrightarrow\ \ &(k+1)(k+2)(k+3)\\ \\ \Longrightarrow\ \ &(k+1)(k+2)k+(k+1)(k+2)3\\ \\ \Longrightarrow\ \ &k(k+1)(k+2)+3(k+1)(k+2)\\ \\ \Longrightarrow\ \ &6m+3(2q)\ \ \ \ \ \ \ \ \ \ [\ \because\ 2 \mid n(n+1)\ ]\\ \\ \Longrightarrow\ \ &6m+6q\\ \\ \Longrightarrow\ \ &6(m+q)\end{aligned}$

Hence Proved!