simple relation to prove!

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.