题目

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. 答案：【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=a1q4+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. He is ill in hospital, __         he can’t come here.    A. so                   B. if                      C. because