(1)设{an}的公比为q(q≠0,q≠1),利用a5,a3,a4成等差数列结合通项公式,可得,由此即可求得数列{an}的公比;
(2)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,从而得证.
(1)【解析】
设{an}的公比为q(q≠0,q≠1)
∵a5,a3,a4成等差数列,∴2a3=a5+a4,
∴
∵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.