Question

Prove the following by using Mathematical Induction by using $n$ in $N$ :
$n\left( n+1 \right)\left( n+5 \right)$ is multiple of $3$ .

Answer 1

To prove this statement using mathematical induction, we need to first find the first two steps of the mathematical induction which is necessary to prove the certain statement.
Step One: First prove that the statement or the equation is true for $n=1$.
Step Two: Assume any variable (in this case) let us presume the variable as $k$, now prove if the following statement or equation is true for $n=k$ and if, true then also true for $n=k+1$.

Complete step-by-step solution:
Now placing the value of $n=1$ in the equation $n\left( n+1 \right)\left( n+5 \right)$ to check whether it’s a multiple of three or not.
For $n=1$, $n\left( n+1 \right)\left( n+5 \right)$
$\Rightarrow 1\left( 1+1 \right)\left( 1+5 \right)$
$\Rightarrow 1\times 2\times 6$
$\Rightarrow 12$
Hence, the value $12$ is a multiple of $3$ as $3\times 4$ thereby, the term $n\left( n+1 \right)\left( n+5 \right)$ is correct for $n=1$.
After successfully completing the first step let us proceed with the second step of the hint by placing $n=k$.
For $n=k$, [\ n \left( n+1 \right)] $\left( n+5 \right)$
$\Rightarrow k\left( k+1 \right)\left( k+5 \right)$
$\Rightarrow k\left( {{k}^{2}}+6k+5 \right)$
$\Rightarrow k\left( {{k}^{2}}+6k+5 \right)$
$\Rightarrow {{k}^{3}}+6{{k}^{2}}+5k$
Now assuming that the term ${{k}^{3}}+6{{k}^{2}}+5k$ is a multiple of $3$ we equate the result to $3r$ where $r$ is the multiple of three which upon multiplication leads to ${{k}^{3}}+6{{k}^{2}}+5k$ as shown below:
$3\times r={{k}^{3}}+6{{k}^{2}}+5k$.
Now let us place the value of $n$ as $n=k+1$
For $n=k+1$, $n\left( n+1 \right)\left( n+5 \right)$
$\Rightarrow k+1\left( \left( k+1 \right)+1 \right)\left( \left( k+1 \right)+5 \right)$
$\Rightarrow \left( {{k}^{2}}+3k+2 \right)\left( k+6 \right)$
$\Rightarrow {{k}^{3}}+9{{k}^{2}}+20k+12$
There are already three variables with different degrees let us replace the value of ${{k}^{3}}$ as ${{k}^{3}}=3r-6{{k}^{2}}-5k$ from $3\times r={{k}^{3}}+6{{k}^{2}}+5k$ in the above equation for $n=k$.
$\Rightarrow {{k}^{3}}+9{{k}^{2}}+20k+12$
$\Rightarrow 3r-6{{k}^{2}}-5k+9{{k}^{2}}+20k+12$
$\Rightarrow 3r+3{{k}^{2}}+15k+12$
$\Rightarrow 3\left( r+{{k}^{2}}+5k+4 \right)$
Now as we can see that the above $3\left( r+{{k}^{2}}+5k+4 \right)$ is a multiple of $3$ as $3\times \left( r+{{k}^{2}}+5k+4 \right)$.
Hence, the term $n\left( n+1 \right)\left( n+5 \right)$ is a multiple of $3$ for $n=k+1$ as it is for $n=k$.
Hence, using mathematical induction we can state that $n\left( n+1 \right)\left( n+5 \right)$ is a multiple of $3$ for all the values of $n\in N$.

Note: Students may go wrong while solving for all the steps required, one must solve for all three values of $n$ as $1,k,k+1$ as it is necessary to show that all three cases give the end result as multiple of $3$. Now the solution for $n=k+1$ should be done carefully as it’s a three variable multiplication so first multiply $k+1\left( \left( k+1 \right)+1 \right)$ of $k+1\left( \left( k+1 \right)+1 \right)\left( \left( k+1 \right)+5 \right)$ and then multiply the rest and upon multiplication carefully replace the value of ${{k}^{3}}$ by its above equation when $n=k$, be careful of the sign when replacing the value of ${{k}^{3}}$ in ${{k}^{3}}+9{{k}^{2}}+20k+12$.