Question

设数列{an}的各项都是正数,a1=1,(an+1)/(an+1+1)=an+1/2an,bn=an2+an.(1)求数列{bn}的通项公式；(2)求

Answer 1

Answer 2

1...
(An+1)/{A(n+1)+1}=A(n+1)/2An
∴{A(n+1)^2+A(n+1)}/{(An)^2+An}=2
∵Bn=(An)^2+An ∴B(n+1)={A(n+1)^2+A(n+1)} ∴B(n+1)/Bn=2 既Bn为公比2的等比数列 B1=(a1)^2+a1=2
所以Bn=2*2^(n-1)=2^n
2...
∵ Bn=(An)^2+An Bn=2^n ∴(An)^2+An=2^2 解得An=√(2^n+1/4)-1/2
3...(1+An)A(n+1)
={√(2^n+1/4)+1/2}*{√(4^n+1/4)-1/2}
>{√(2^n+1/4)+1/2}*{√(2^n+1/4)-1/2}
=2^n

∴1/(1+An)A(n+1)<1/2^n
∴原式
< 1/2^1+1/2^2+……+1/2^n
设原式=Cn=1/2^1+1/2^2+……+1/2^n
则2Cn=1+1/2^1+1/2^2+……+1/2^(n-1)
2Cn-Cn=1-1/2^n
即Cn=1-1/2^n<1

∴原式<1