Prove by induction that 3n(n 1) is divisible by 6 for all positive integers.

Answer with explanation:We are asked to prove by the method of mathematical induction that:       3n(n+1) is divisible by 6 for all positive integers.for n=1 we have:[tex]3n(n+1)=3\times 1(1+1)\\\\i.e.\\\\3n(n+1)=3\times 2\\\\i.e.\\\\3n(n+1)=6[/tex]which is divisible by 6.Hence, the result is true for n=1Let the result is true for n=ki.e. 3k(k+1) is divisible by 6.Now we prove that the result is true for n=k+1Let n=k+1then[tex]3n(n+1)=3(k+1)\times (k+1+1)\\\\i.e.\\\\3n(n+1)=3(k+1)(k+2)\\\\i.e.\\\\3n(n+1)=(3k+3)(k+2)\\\\i.e.\\\\3n(n+1)=3k(k+2)+3(k+2)\\\\i.e.\\\\3n(n+1)=3k^2+6k+3k+6\\\\i.e.\\\\3n(n+1)=3k^2+3k+6k+6\\\\i.e.\\\\3n(n+1)=3k(k+1)+6(k+1)[/tex]Since, the first term:[tex]3k(k+1)[/tex] is divisible by 6.( As the result is true for n=k)and the second term [tex]6(k+1)[/tex] is also divisible by 6.Hence, the sum:[tex]3k(k+1)+6(k+1)[/tex]  is divisible by 6.Hence, the result is true for n=k+1Hence, we may say that the result is true for all n where n belongs to positive integers.