问题
解答题
已知数列{xn}中,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,
(1)求数列{yn}的通项公式;
(2)证明:数列{xn}为等比数列;
(3)设数列{xn}的前n项和为Sn,若对一切正整数n,Sn<a恒成立,求实数a的取值范围.
答案
(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=
=128(1-26(1-
)1 2n 1- 1 2
)<1281 2n
Sn=128,lim n→∞
故所求a的取值范围为a≥128.