In the expansion of (1+x)^n, the successive coefficients are a₀, a₁, a₂, a₃,......,aₙ; show that a₀+ 2a₁+ 3a₂+......+(n+1)aₙ = 2^n + n*2^(n-1)

let (1+x)^n= a0 + a1*x^1 + a2*x^2 + ….... + an*x^n........(1)

multiply by x on both sides

x(1+x)^n= a0*x^1 + a1*x^2 + a2*x^3 + ….... + an*x^(n+1)

differentiate both sides wrt x

(1+x)^n + nx(1+x)^(n-1) = a0 + 2a1*x^1 + 3a2*x^2 + …..... + (n+1)an*x^n

put x=1, we get

hence proved.

Given, (1+x)^n.(1+y)^n.(1+z)^n
 
=[(1+x)(1+x)...(1+x)]n factors
=(1+y)(1+y)+....(1+y)]n factors
=[(1+z)(1+z)...(1+)z]n factors
There are 3n factors in (1+x)^n.(1+y)^n.(1+z)^n
  for term of degree r. 
We choose r brackets out of 3n brackets and then multiply second terms in each bracket.
There are 3nCr such terms each having coefficient 1. 
Hence, the sum of the coefficients is  
3nCr.