证：由数列{an}是等差数列,得an=a1+(n-1)d ,其中a1为首项,d为公差.b1b2b3=[(1/2)^(a1)][(1/2)^(a1+d)][(1/2)^(a1+2d)]=(1/2)(a1+a1+d+a1+2d)=(1/2)(3a1+3d)=[(1/2)³]^(a1+d)=(1/8)^(a2)=1/8a2=1b1+b2+b3=(1/2)^...