Question

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

Answer 1

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 3
We 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 3
So $P(k + 1)$ is also true.
Since $P(k)$ is true whenever $P(k)$ is true.
Hence proved $n(n + 1)(n + 5)$ is a multiple of 3
So, 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 $n$.