数列证明题一题设数列{An}满足:A1=1,且当n∈N*时,An^3+An^2×[1-A(n+1)]+1=A(n+1)求
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数列证明题一题
设数列{An}满足:A1=1,且当n∈N*时,
An^3+An^2×[1-A(n+1)]+1=A(n+1)
求证:数列{An}是递增数列.
设数列{An}满足:A1=1,且当n∈N*时,
An^3+An^2×[1-A(n+1)]+1=A(n+1)
求证:数列{An}是递增数列.
证明:由题
A(n+1)=[An^3+An^2+1]/[An^2+1]
=[An^3+An+An^2+1-An]/[An^2+1]
=An+1-An/[An^2+1]
故A(n+1)-A(n)
=1-An/[An^2+1]
=[An^2+1-An]/[An^2+1]
=[(An-1/2)^2+3/4]/[An^2+1]
>=0
故A(n+1)>=A(n)
故数列{An}是递增数列
A(n+1)=[An^3+An^2+1]/[An^2+1]
=[An^3+An+An^2+1-An]/[An^2+1]
=An+1-An/[An^2+1]
故A(n+1)-A(n)
=1-An/[An^2+1]
=[An^2+1-An]/[An^2+1]
=[(An-1/2)^2+3/4]/[An^2+1]
>=0
故A(n+1)>=A(n)
故数列{An}是递增数列
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