Let P(n) denotes that \(\displaystyle{n}{\left({n}^{{{n}+{5}}}\right)}\) is divisible by 6, for each natural number n.

\(\displaystyle{P}{\left({1}\right)}:{1}{\left({1}^{{{2}+{5}}}\right)}={6}\), which is divisible by 6. Therefore the result is true for n=1.

Now, assume that P(n) is true for n=k. That is \(\displaystyle{P}{\left({k}\right)}:{k}{\left({k}^{{{2}+{5}}}\right)}\text{ is }\div{i}\text{ is }\le\text{ by }{6}.{k}{\left({k}^{{{2}}}+{5}\right)}={6}{m}\) for some m∈N. We have to prove that P(k+1) is also true.

\(P(k+1):(k+1)[(k+1)^{2+5}] =(k+1)[k^{2}+2k+1+5] =(k+1)[k^{2}+2k+6] =k^{3}+2k^{2}+6k+k^{2}*2k+6 =k^{3}+3k^{2}+8k+6 =k^{3}+5k+3k^{2}+3k+6 =k(k^{2}+5)+3(k^{2}+k+2) =(6m)+3(k^{2)+k+2)}\)

Since \(\displaystyle{k}^{{{2}}}+{k}+{2}\) is divisible by 22 for every even and odd k. Therefore \(\displaystyle{3}{\left({k}^{{{2}}}+{k}+{2}\right)}\) is divisible by 66 and hence \(\displaystyle{\left({6}{m}\right)}+{3}{\left({k}^{{{2}}}+{k}+{2}\right)}\) is divisible by 6. Therefore, \(\displaystyle{P}{\left({k}+{1}\right)}\) is true whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true.