let p(n) denote the above express.
put n=1;
1*2*3=6 is divisible by 6.
let p(k) be true;
k*(k+1)*(2k+1)=6U(u here is any constant)
2(k^3)+3(k^2)+k=6u
prove that p(k+1) is true
(k+1)*(k+2)*(2k+3);
= 2(k^3)+9(k^2)+13k+6
=6u+6(k^2)+12k+6
=6(u+(k^2)+2k+1)
is divisible by 6
hence p(n) is true for all natural numbers i.e n>0.