一道数列的题目,很紧急.设{an}是等差数列,bn=(1/2)^an,已知b1+b2+b3=21/8,b1 x b2 x

设{An}的公差是d,第二项是a
A1=a-d,A2=a,A2=a+d
(1/2)^(a-d)+(1/2)^a+(1/2)^(a+d)=21/8
[(1/2)^(a-d)][(1/2)^a][(1/2)^(a+d)]=1/8
化简第二式得,(1/2)^(3a)=1/8=(1/2)^3,求得a=1
将a的值代入第一式
(2^d)/2+1/2+(1/2)/(2^d)=21/8
令t=2^d>0
t/2+1/2+1/(2t)=21/8
t²+t+1=21t/4
4t²-17t+4=0
(4t-1)(t-4)=0
t=1/4或t=4
2^d=1/4或2^d=4
d=-2或t=2
{An}的首项是3,公差是-2,或首项是-1,公差是2
通项公式是：An=-2n+5或An=2n-3

设an=a1+(n-1)d
a1=a2-d a3=a2+d
bn=(1/2)^an
b1+b2+b3=(1/2)^a1+(1/2)^a2+(1/2)^a3
=(1/2)^(a2-d)+(1/2)^a2+(1/2)^(a2+d)=21/8
b1b2b3=(1/2)^(a1+a2+a3)=(1/2)^3a2=1/8
a2=1
代入化简 得 (1/2)^(1-d)+(1/2)^(1+d)=17/8
2^(1+d)+2^(1-d)=17/2
d=2
a1=1-2=-1
an=-1+2(n-1)=2n-3