Prove that `1*2+2*3+3*4+.....+n*(n+1)=(n(n+1)(n+2))/(3)`

Explanation

Let P(n) : `1*2+2*3+3*4+.....+n*(n+1)=(n(n+1)(n+2))/(3)` be the given statement
Step 1 : Put n=1
Then `LHS=1*2=2`
`RHS=(1(1+2)(1+2))/(3)=(2xx3)/(3)=2`
`:. LHS=RHS`
`rArrP(n)` is true for n=1
Step 2: Assume that P(n) is true for n=k
`:.1+*2+2*3+3*4+....+k(k+1)=(k(k+1)(k+2))/(3)`
Adding `(k+1)(k+2)` on both sides, we get
`1*2+2*3+3*4+.....k(k+1)+(k+1)(k+2)` ltDbrgt `=(k(k+1)(k+2))/(3)+(k+1)(k+2)=(k+1)(k+2)((k)/(3)+1)=((k+1)(k+2)(k+3))/(3)`
`:.1*2+2*3+3*4+....+k*(k+1)(k+1)(k+2)=((k+1)(bar(k+1)+1)(bar(k+1)+2))/(3)`
`rArrP(n)` is true for `n=k+1`
`:.` By the principle of matheatical induction o P(n) is true for all natural numbrs
Hence, `1*2+2*3+3*4+.....+n*(n+1)=(n(n+1)(n+2))/(3), n in N`