已知等比数列{an}的公比q= -1/2 (2)证明 对任意k∈N*,ak,ak+2,ak+1成等差数列

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已知等比数列{an}的公比q= -1/2 (2)证明 对任意k∈N*,ak,ak+2,ak+1成等差数列

已知等比数列{an}的公比q= -1/2 (2)证明 对任意k∈N*,ak,ak+2,ak+1成等差数列
已知等比数列{an}的公比q= -1/2 (2)证明 对任意k∈N*,ak,ak+2,ak+1成等差数列

已知等比数列{an}的公比q= -1/2 (2)证明 对任意k∈N*,ak,ak+2,ak+1成等差数列
你的k+2、k+1是下标吧
证明:
ak+2=ak*q²
ak+1=ak*q
ak+(ak+1)=ak+ak*q=ak(1+q)=1/2ak
ak+2=ak*q²=1/4ak
∵ak+(ak+1)=2(ak+2)
∴ak,ak+2,ak+1成等差数列

设an=a1*(-1/2)^(n-1)=-2a1*(-1/2)^n
因为是等差数列
所以 2a(k+2)=ak+a(k+1)
带入得 2*(-2)*a1(-1/2)^(k+2)=(-2)*a1(-1/2)^k+(-2)*a1*(-1/2)^(k+1)
整理得到 2*(-1/2)^(k+2)=(-1/2)^(k)+(-1/2)^(k+1) 约去(-1/2)^(k)
得1/2=1-1/2
所以得证