Prove that n(n+1)(2n+1) is divisible by 6 for all n?N by using principle of mathematical induction.

Dear student,

Let the given statement be $P\left(n\right)=n\left(n+1\right)\left(2n+1\right)$ which is divisible by 6
For $n=1, P\left(1\right): 1\times \left(1+1\right)\left(2+1\right)=6$,  which is true
Assume that $P\left(k\right)$ is true for some positive integers k  i.e., P(k) is divisible by 6
We shall now prove that P(k + 1) is also true
                  $$\begin{gathered} P(k+1)=\left(k+1\right)\left(k+2\right)\left[2\left(k+1\right)+1\right]=\left(k+1\right)\left(k+2\right)\left[2k+3\right]=\left(k+1\right)\left(k+2\right)\left[2k+1+2\right] \\ =k\left(k+1\right)\left[\left(2k+1\right)+2\right]+2\left(k+1\right)\left[\left(2k+1\right)+2\right] \\ =k(k+1)(2k+1)+2k(k+1)+2(k+1)(2k+1)+4(k+1) \\ =k(k+1)(2k+1)+2(k+1)\left[k+2k+1+2\right] \\ =k(k+1)(2k+1)+2(k+1)\left[3k+3\right]=k(k+1)(2k+1)+6(k+1)\left(k+1\right) \end{gathered}$$
First term is divisible by 6 as we assumed, and second term is also divisible by 6.
Thus, P(k + 1) is divisible by 6, whenever P(k) is true.
Hence, from the principle of mathematical induction P(n) is true for $n\in N$

Regards