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**Hint:**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 \[{{k}^{3}}+9{{k}^{2}}+20k+12\].

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