For all n∈N,$$n(n+1)(n+5)$$ is a multiple of

Question

For all  n∈N,$$n(n+1)(n+5)$$ is a multiple of

Infinity Learn · Accepted Answer

Let the given statement be $$P(n)$$.  i.e.  $$P(n)$$:$$n(n+1)(n+5)$$ is a multiple of $$3$$.Step l: For $$n=1$$,$$1(1+1)(1+5)=1$$×$$2$$×$$6=12=3$$×$$4which$$ is a multiple of $$3$$, that is trueStep lI: Let it is true for n $$=$$ k, $$k(k+1)(k+5)=3$$λ⇒$$kk2+5k+k+5=3$$λ⇒$$k3+6k2+5k=3$$λ$$----iStep$$ III: For $$n=k+1$$, (k+1)(k+1+1)(k+1+5)=(k+1)(k+2)(k+6)=k2+2k+k+2(k+6)=k2+3k+2(k+6)=k3+6k2+3k2+18k+2k+12=k3+9k2+20k+12=3$$λ−$$6k2$$−$$5k+9k2+20k+12$$  [using Eq. $$(i) $$=3$$λ$$+3k2+15k+12=3$$λ$$+k2+5k+4which$$ is a multiple of $$3$$. Therefore, $$P(k$$ $$+$$ $$1)$$ is true when $$P(k)$$ is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.

For all $n \in N, n (n + 1) (n + 5)$ is a multiple of

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For all n∈N,n(n+1)(n+5) is a multiple of

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For all $n \in N, n (n + 1) (n + 5)$ is a multiple of