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

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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
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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
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

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
2a(k+2)=2a1·q^(k+1)=2a1·(-1/2)^(k+1)=-a1·(-1/2)^k
ak+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)^k
2a(k+2)=ak+a(k+1)
数列{an}是等差数列.