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](/uploads/image/z/17829133-61-3.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%85%AC%E6%AF%94q%3D+-1%2F2+%E8%AF%81%E6%98%8E+%E5%AF%B9%E4%BB%BB%E6%84%8Fk%E2%88%88N%2A%2Cak%2Cak%2B2%2Cak%2B1%2C%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
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.
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.
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