已知等比数列{an}的公比q= -1/2 证明 对任意k∈N*,ak,ak+2,ak+1,成等差数列
问题描述：
已知等比数列{an}的公比q= -1/2 证明 对任意k∈N*,ak,ak+2,ak+1,成等差数列
答
2a(k+2)=2a1·q^(k+1)=2a1·(-1/2)^(k+1)=-a1·(-1/2)^kak+a(k+1)=a1·q^(k-1)+a1·q^k=a1·[(-1/2)^(k-1)+(-1/2)^k]=a1·(-1/2)^k ·[(-2)+1]=-a1·(-1/2)^k2a(k+2)=ak+a(k+1)数列{an}是等差数列.