数列{an} {bn}满足:a1=1,2an+bn=1,a(n+1)=an(bn+1).(1)求an的表达式.(2)若cn=ana(n+1),求数列{cn}的前n项和

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数列{an} {bn}满足:a1=1,2an+bn=1,a(n+1)=an(bn+1).(1)求an的表达式.(2)若cn=ana(n+1),求数列{cn}的前n项和

数列{an} {bn}满足:a1=1,2an+bn=1,a(n+1)=an(bn+1).(1)求an的表达式.(2)若cn=ana(n+1),求数列{cn}的前n项和
数列{an} {bn}满足:a1=1,2an+bn=1,a(n+1)=an(bn+1).(1)求an的表达式
.(2)若cn=ana(n+1),求数列{cn}的前n项和

数列{an} {bn}满足:a1=1,2an+bn=1,a(n+1)=an(bn+1).(1)求an的表达式.(2)若cn=ana(n+1),求数列{cn}的前n项和
由2an+bn=1,a(n+1)=an(bn+1)
=> a(n+1)=an(2-2an) =>a(n+1)=-2an^2+2an=-2(an-1/2)^2+1/2
=>a(n+1)-1/2=-2(an-1/2)^2
题似乎有问题吧,倒这一步an+1就等于0了

n=1时,将a1=1带入后两个关系式,约去bn,求出a2,再用a2求a3,数学归纳法求an

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