设存在an=a1+(n-1)d
a1C(n,0)+a2C(n,1)+a3C(n,2)+…+a（n+1）C(n,n)
=a1C(n,0)+(a1+d)C(n,1)+(a1+2d)C(n,2)+…+[a1+(n-1)d]C(n,n-1)+[a1+nd]C(n,n)
=a1[C(n,0)+C(n,1)+C(n,2)+…+C(n,n)]+[dC(n,1)+2dC(n,2)+…+(n-1)dC(n,n-1)+ndC(n,n)]
=a1*2^n+d[C(n,1)+2C(n,2)+…+(n-1)C(n,n-1)+nC(n,n)]
=a1*2^n+(1/2)d{nC(n,n)+[C(n,1)+(n-1)C(n,n-1)]+[2C(n,2)+(n-2)C(n-2)]+…+[(n-2)C(n,n-2)+2C(n,2)]+[(n-1)C(n,n-1)+C(n,1)]+nC(n,n)}
=a1*2^n+(1/2)d{nC(n,0)+nC(n,1)+nC(n,2)+…+nC(n,n-2)+nC(n,n-1)+nC(n,n)}
=a1*2^n+(1/2)dn{C(n,0)+C(n,1)+C(n,2)+…+C(n,n-2)+C(n,n-1)+C(n,n)}
=a1*2^n+(1/2)dn*2^n
=[a1+(1/2)dn]2^n
=n*2^n
要使上式恒成立,只要a1+(1/2)dn=n恒成立,
只要a1=n(1-d/2)恒成立,
所以当a1=0,d=2时可满足要求,
所以
an=2n-2为所求.