By using the principle of mathematical induction,Prove that \[n(n + 1)(n + 5)\] is a multiple of 3.

Question

By using the principle of mathematical induction,Prove that $$n(n + 1)(n + 5)$$ is a multiple of 3.

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Hint:In this question we have to prove that the $$n(n + 1)(n + 5)$$ is a multiple of 3. Let assume $$P(n) = n(n + 1)(n + 5)$$. Hence we have to prove that  $$P(n) = n(n + 1)(n + 5) = 3d$$ where $$d \in N$$ by using the principle of mathematical induction.Complete step-by-step answer:We have to prove that $$n(n + 1)(n + 5)$$ is a multiple of 3We will prove it by using the concept of mathematical induction method for all $$n \in N$$Let $$P(n) = n(n + 1)(n + 5) = 3d$$ where $$d \in N$$For $$n = 1$$Since $$P(n) = n(n + 1)(n + 5)$$Substituting the value $$n = 1$$in above equation,$$P(1) = 1(1 + 1)(1 + 5)$$$$P(1) = 1(2)(6) = 12$$ which is divisible by 3$$P(n)$$ is true for $$n = 1$$By using the induction method So, $$P(k)$$ is also true Again, we check for $$P(k)$$We can write $$P(k) = k(k + 1)(k + 5) = 3m$$ where $$m \in N$$Simplify and we get$$ \Rightarrow {k^3} + 6{k^2} + 5k = 3m$$$$ \Rightarrow {k^3} =  - 6{k^2} - 5k + 3m$$Now we will prove that $$P(k + 1)$$ is true so in the place of $$k$$ replace $$k + 1$$$$P(k + 1) = (k + 1)(k + 2)(k + 6) = {k^3} + 9{k^2} + 20k + 12\;$$Putting the value of $${k^3}$$ in above equation we get,$$ = (3m - 6{k^2} - 5k) + 9{k^2} + 20k + 12$$Adding the terms have same power we get,$$ = 3m + 3{k^2} + 15k + 12$$Taking 3 as a common term,$$ = 3(m + {k^2} + 5k + 4)$$We can write $$3r$$, where $$r = m + {k^2} + 5k + 4$$We can see that $$P(k + 1)$$is multiply by 3So $$P(k + 1)$$ is also true.Since $$P(k)$$ is true whenever $$P(k)$$ is true.Hence proved $$n(n + 1)(n + 5)$$ is a multiple of 3So, by the principle of induction, $$P(n)$$ is divisible by 3 for all $$n \in N$$When a statement is true for a natural number $$n = k$$, then it will also be true for its successor, $$n = k + 1$$, and the statement is true for $$n = 1$$, then the statement will be true for every natural number $$n$$.Note:Mathematical induction: A proof by induction consists of two cases. The first, the base case (or basis), proves the statement for $$n = \;1$$ without assuming any knowledge of other cases. The second case, the induction step, proves that if the statement holds for any given case $$n = k$$, then it must also hold for the next case $$n\; = \;k\; + {\text{ }}1$$. These two steps establish that the statement holds for every natural number $$n$$.