n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.
Let P(n) be the given statement.
Now,
$P(n): n(n+1)(n+5)$ is a multiple of 3 .
Step 1:
$P(1): 1(1+1)(1+5)=12$
It is a multiple of 3 .
Hence, $P(1)$ is true.
Step 2 :
Let $P(m)$ be true.
Then, $m(m+1)(m+5)$ is a multiple of 3 .
Suppose $m(m+1)(m+5)=3 \lambda$, where $\lambda \in N$.
We have to show that $P(m+1)$ is true whenever $P(m)$ is true.
Now,
$P(m+1)=(m+1)(m+2)(m+6)$
$=m(m+1)(m+6)+2(m+1)(m+6)$
$=m(m+1)(m+5+1)+2(m+1)(m+6)$
$=m(m+1)(m+5)+m(m+1)+2(m+1)(m+6)$
$=3 \lambda+(m+1)(m+2 m+6)$$[$ From $P(m)]$
$=3 \lambda+3(m+1)(m+2)$
It is clearly a multiple of 3 .
Thus, $P(m+1)$ is true.
By the principle of mathematical $i$ nduction, $P(n)$ is true for all $n \in N$.
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