# Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ :Q: 19        $n(n+1)(n+5)$  is a multiple of $3.$

Let the given statement be p(n) i.e.
$p(n):n(n+1)(n+5)$
For n = 1  we have
$p(1):1(1+1)(1+5)=1.2.6=12$   ,   which is multiple of 3, hence true

For  n = k  we have

$p(k):k(k+1)(k+5) \ \ \ \ \ \ \ -(i)$   ,        Let's assume that this is multiple of 3 = 3m

Now,
For  n = k + 1  we have
$p(k+1):(k+1)((k+1)+1)((k+1)+5)$                                                                                                                                                                                                                                                          $=(k+1)(k+2)((k+5)+1)$

$=(k+1)(k+2)(k+5)+(k+1)(k+2)$
$=k(k+1)(k+5)+2(k+1)(k+5)+(k+1)(k+2)$
$=3m+2(k+1)(k+5)+(k+1)(k+2) \ \ \ \ \ \ \ \ (using \ (i))$
$=3m+(k+1)(2(k+5)+(k+2)) \ \ \ \ \ \ \ \ (using \ (i))$
$=3m+(k+1)(2k+10+k+2)$
$=3m+(k+1)(3k+12)$
$=3m+3(k+1)(k+4)$
$=3(m+(k+1)(k+4) )$
$=3l$                                                Where $\left ( l=(m+(k+1)(k+4) ) \right )$   some natural number

Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is multiple of 3 for all natural numbers n

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