We have, (a+b+c)n=[a+(b+c)]n =an+nC1an−1(b+c)1+nC2an−2(b+c)2+... ...+nCn(b+c)n
Further, expanding each term of R.H.S., we note that first term consist of 1 term. Second term on simplification gives 2 terms. Third term on simplification gives 3 terms and so on. ∴ The total number of terms =1+2+3+...+(n+1) =2(n+1)(n+2)