find the sum to n terms of the given series 1*2*3+2*3*4+3*4*5+....?

Question

Please answer with details.

Answer 1

=n*(n+1)*(n^2+5n+6)/4
sum(i*(i+1)*(i+2)) for i=1 till i=n
=sum(i^3+3i^2+2i)
sum(i)=n*(n+1)/2
sum(i^2)=n*(n+1)*(2n+1)/6
sum(i^3)=(n*(n+1)/2)^2
==>
sum(i^3+3i^2+2i)
=(n*(n+1)/2)^2
+3n*(n+1)*(2n+1)/6
+2*n*(n+1)/2
=n*(n+1)*(n^2+5n+6)/4
Sorry for my English