设数列{An}和{bn}满足A1=1/2,2nA(n+1)=(n+1)An,且Bn=ln(1+An)+1/2(An)2,
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设数列{An}和{bn}满足A1=1/2,2nA(n+1)=(n+1)An,且Bn=ln(1+An)+1/2(An)2,n属于N+(1):求A2,A3,A4,并求数列{An}的通项公式(2):对一切n属于N+,证明2/(An+2)小于An/Bn成立
(1)A1=1/2,2nA(n+1)=(n+1)An,
∴A/(n+1)=(1/2)An/n=…=1/2^n,
∴An=n/2^n.A2=1/2,A3=3/8,A4=1/4.
(2)An(An+2)-2Bn
=2[An-ln(1+An)]>0,
An,Bn>0,
∴原式成立.
∴A/(n+1)=(1/2)An/n=…=1/2^n,
∴An=n/2^n.A2=1/2,A3=3/8,A4=1/4.
(2)An(An+2)-2Bn
=2[An-ln(1+An)]>0,
An,Bn>0,
∴原式成立.
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