设数列{an}是等差数列,bn=(1/2)的an次方,又b1+b2+b3=21/8,b1b2b3=1/8,证明数列｛bn

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)=1/8
a2=1
b1+b2+b3=(1/2)^(a1)+(1/2)^(a2)+(1/2)^(a3)=21/8
a2=1代入
(1/2)^(1-d)+(1/2)+(1/2)^(1+d)=21/8
整理,得
4[(1/2)^d]² -17(1/2)^d +4=0
[4(1/2)^d -1][(1/2)^d -4]=0
(1/2)^d=1/4或(1/2)^d=4
d=2或d=-2
(1)
d=2时,a1=a2-d=1-2=-1 an=a1+(n-1)d=-1+2(n-1)=2n+1
bn=(1/2)^(2n+1)
b1=(1/2)^(-1)=2
b(n+1)/bn=(1/2)^[2(n+1)+1]/(1/2)^(2n+1)=1/4,为定值.
数列{bn}是以2为首项,1/4为公比的等比数列.
(2)
d=-2时,a1=a2-d=1-(-2)=3 an=a1+(n-1)d=3-2(n-1)=5-2n
bn=(1/2)^(5-2n)
b1=(1/2)^(5-2)=1/8
b(n+1)/bn=(1/2)^[5-2(n+1)]/(1/2)^(5-2n)=4,为定值.
数列{bn}是以1/8为首项,4为公比的等比数列.
综上,得{bn}是等比数列.