解题思路:(1)由x1,x5是方程log22x-8log2x+12=0的两根,等差数列{yn}满足yn=log2xn,且其公差为负数,能够推导出y1=log2x1=6,y5=log2x5=2,yn=7-n.
(2)由yn=log2xn=7-n,yn+1=log2xn+1=6-n,知
x
n+1
x
n
=
2
6−n
2
7−n
=
1
2
,由此能够证明数列{xn}为等比数列.
(3)
S
n
=
2
6
(1−
1
2
n
)
1−
1
2
=128(1−
1
2
n
)<128
lim
n→∞
S
n
=128
,由此能求出a的取值范围.
(1)∵x1,x5是方程log22x-8log2x+12=0的两根,
∴log2x1+log2x5=8,log2x1•log2x5=12,
∵等差数列{yn}满足yn=log2xn,且其公差为负数,
∴log2x1=6,log2x5=2.
y1=log2x1=6,y5=log2x5=2,yn=7-n.
(2)∵yn=log2xn=7-n,yn+1=log2xn+1=6-n
∴
xn+1
xn=
26−n
27−n=
1
2,
∴数列{xn}为等比数列.
(3)Sn=
26(1−
1
2n)
1−
1
2=128(1−
1
2n)<128
lim
n→∞Sn=128,
故所求a的取值范围为a≥128.
点评:
本题考点: 数列与不等式的综合;根与系数的关系;函数恒成立问题;等差数列的通项公式;等比关系的确定.
考点点评: 本题考查通项公式的求法、等比数列的证明和实数a的取值的求法,解题时要认真审题,注意挖掘题设中的隐含条件,合理地进行等价转化.