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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.
题目详情
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.
(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.
▼优质解答
答案和解析
(1)设{an}的公比为q(q≠0,q≠1)
∵a5,a3,a4成等差数列,∴2a3=a5+a4,
∴2a1q2=a1q4+a1q3
∵a1≠0,q≠0,
∴q2+q-2=0,解得q=1或q=-2
∵q≠1,
∴q=-2
(2)证明:对任意k∈N+,Sk+2+Sk+1-2Sk=(Sk+2-Sk)+(Sk+1-Sk)=ak+2+ak+1+ak+1=2ak+1+ak+1×(-2)=0
∴对任意k∈N+,Sk+2,Sk,Sk+1成等差数列.
∵a5,a3,a4成等差数列,∴2a3=a5+a4,
∴2a1q2=a1q4+a1q3
∵a1≠0,q≠0,
∴q2+q-2=0,解得q=1或q=-2
∵q≠1,
∴q=-2
(2)证明:对任意k∈N+,Sk+2+Sk+1-2Sk=(Sk+2-Sk)+(Sk+1-Sk)=ak+2+ak+1+ak+1=2ak+1+ak+1×(-2)=0
∴对任意k∈N+,Sk+2,Sk,Sk+1成等差数列.
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