Series nC1+1⋅nC2+2⋅nC3+…+n⋅nCn is equal to

Since, (1+x)n=nC0+nC1x+nCx2+…+nCnxn  Ondifferentiatingw.r.t., x, weget n(1+x)n−1=0+nC1+2⋅x⋅nC2+…+n⋅nCnxn−1Put x=1, we get n(1+1)n−1=nC1+2⋅nC2+…+n⋅nCn⇒  n2n−1=nC1+2⋅nC2+…+n⋅nCn