Question

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

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right)=n(n+1)(n+5)$ is a multiple of $3$.

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right)=1(1+1)(1+5)$ is a multiple of $3$
$\Rightarrow 12$ is multiple of $3$ (which is true)
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$.
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(k\right):K(K+1)(K+5)$ is a multiple of $3$
$\Rightarrow K(K+1)(K+5)=3m$, where $m\in N\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now we have
$(K+1)\left\{\right(K+1)+1\}\left\{\right(K+1)+5\}$
$=(K+1)(K+2)\left\{\right(K+5)+1\}$
$=(K+1)(K+2)(K+5)+(K+1)(K+2)$
$=K(K+1)(K+5)+2(K+1)(K+5)+(K+1)(K+2)$
Now, using $\left(1\right)$
$=3m+(K+1)\left\{2\right(K+5)+(K+2\left)\right\}$
$=3m+(K+1)\{3K+12\}$
$=3m+3(K+1)(K+4)$
$=3\{m+(K+1\left)\right(K+4\left)\right\}$
$=3C\{\text{where}C=\{m+(K+1)(K+4)\};C\in N$ i.e.,we can say $(K+1)\left\{\right(K+1)+1\}\{K+1\}+5\}$ is multiple of $3$.
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.

Final answer :
Therefore, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all $n\in N$.