设{an}是公比不为1的等比数列,其前n项和为Sn,且a5,a3,a4成等差数列.
(1)求
数列{an}的公比.
(2)证明:对任意k∈N*,Sk+2,Sk,Sk+1成等差数列.
设{an}是公比不为1的等比数列,其前n项和为Sn,且a5,a3,a4成等差数列.
(1)求数列{an}的公比.
(2)证明:对任意k∈N*,Sk+2,Sk,Sk+1成等差数列.
【解析】(1)设数列{an}的公比为q(q≠0,q≠1),
由a5,a3,a4成等差数列,得2a3=a5+a4,即2a1q2=a
1q4+a1q3,
由a1≠0,q≠0得q2+q-2=0,解得q1=-2,q2=1(舍去),
所以q=-2.
(2)对任意k∈N*,
Sk+2+Sk+1-2Sk=(Sk+2-Sk)+(Sk+1-Sk)
=ak+1+ak+2+ak+1=2ak+1+ak+1·(-2)=0,
所以对任意k∈N*,Sk+2,Sk,Sk+1成等差数列.