`1*2*3+2*3*4+...+n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/4 forall n in N.`

Explanation

Let `P(n):1.2.3+2.3.4+.....+n(n+1)(n+2)=(n(n+1)(n+2)(n+3))/(4)` ........(i)
Step I For `n =1, LHS of Eq. (i) `=1.2.3=6`and RHS of Eq. (i) `=(1.(1+1(1+2)(1+3))/(4)=6`
`therefore` LHS = RHS
Therefore , P(1) is true , then
Step II Assume that P(k) is true , then
`P(k):1.2.3+2.3.4+.....+k(k+1)(k+2)=(k(k+1)(k+2)(k+3))/(4)`
Step III For `n=k+1`
`P(k+1):1.2.3+2.3.4+.....+k(k+1)(k+2)+(k+1)(k+2)(k+3)`
`=((k+1)(k+2)(k+3)(k+4))/(4)`
`therefore LHS = 1.2.3+2.3.4+...+k(k+1)(k+2)+(k+1)(k+2)(k+3)`
=(k(k+1)(k+2)(k+3))/(4)+(k+1)(k+2)(k+3)` [by assumption step]
`=((k+1)(k+2)(k+3))/(4)(k+4)`
`=((k+1)(k+2)(k+3)(k+4))/(4)=RHS`
Therefore , `P(k+1)` is true . Hence , by the principle of mathematical induction P(n) is true for all `n epsi N`.