Prove by the principle of mathematical induction, that
`3xx6+6xx9+9xx12+....+(3n)xx(3n+3)=3n(n+1)(n+2)`

Explanation

Let P(n), `3xx6+6xx9+9xx12+....+(3n)xx(3n+3)=3n(n+1)(n+2)`
When n=1
L.H.S.=`3xx6=18`
and R.H.S =`(3xx1)(1+1)(1+2)`
`=3xx2xx3=18`.
`therefore LHS=RHS`. Hence P(1) is true. (a)
Let P(k) be true.
`Rightarrow=3xx6+6xx9xx+9xx12+....(3k)xx(3k+3)`
=3k(k+1)(k+2).....(i)
To prove that P(k+1) be true i.e. `3xx6+6xx9+9xx12...+3kxx(3k+3)+(3k+3)(3k+6)`
=3(k+1)(k+2)(k+3).....(ii)
Adding (3k+3) (3k+6) to both sides of (i), we get
`[3xx6+6xx9+9xx12+.......+3kxx(3k+3)]+(3k+3)(3k+6)`
`=(k+1)(k+2)[3k+9]`
`=3(k+1)(k+2)(k+3)`
Hence P(k+1) is true whenever P(k) to true. ....(b)
From (a) and (b) by the principal of mathematical induction it follows that P(n) is true for all natural number n.