Question

f(x)=x^2+x,数列An的首项a1>0,a(n+1)=f(an)

若a1=1/2,求1<1/(1+a1)+1/(1+a2)+.................+1/(1+an)<2,(n>=2)
求求你帮帮忙啊！！！！！！！

Answer 1

a(n+1)=an^2+an=an*(an+1)

1/(an+1)=1/an-1/a(n+1)

1/(1+a1)+1/(1+a2)+.................+1/(1+an)
=1/a1-1/a(n+1)
=2-1/a(n+1)<2

a(n+1)-an=an^2>=0

an>0恒成立

1/(1+a1)+1/(1+a2)+.................+1/(1+an)>1/(a1+1)=1

得证

追问

不过 1/(a1+1)不是等于1啊..

追答

我笔误了

1/(1+a1)+1/(1+a2)+.................+1/(1+an)>=1/(a1+1)+1/(a2+1)=1/(1+1/2)+1/(1+3/4)=26/21>1

OK

Answer 2

参考！