高中数学-----------------数列与不等式结合的题目

A(n+1)=1/(2-An)可以用特征根发求通项公式,实在不会的话就用数学归纳法.
本题对应的特征方程为x(2-x)=1所以x=1
所以an=1+1/Cn其中Cn=-2-(n-1)=-n-1所以an=n/(n+1)
bn=1/n第二问比较简单
设Dn=B3n－Bn=1/(n+1)+1/(n+2)+1/(n+3)+·····+1/(n+2n)
所以Dn+1=B3(n+1)－Bn+1=1/(n+2)+1/(n+3)+·····+1/(n+1+2n+2)
所以Dn+1-Dn=1/(n+1+2n)+1/(n+1+2n+1)+1/(n+1+2n+2)-1/(n+1)
=1/(3n+1)+1/(3n+2)+1/(3n+3)-3/(3n+3)
=1/(3n+1)+1/(3n+2)-2/(3n+3)>0
所以Dn+1>Dn即Dn递增,所以D1=B3－B1=1/2+1/3=5/6最小
对任意n∈N+且n≥2都有B3n－Bn＞m/20成立
所以5/6>m/20即m

1------------
A(n+1)
=(1-0*An) /(2- An)
=(2-1*A(n-1))/(3-2*A(n-1))
=(3-2*A(n-2))/(4-3*A(n-2))
=......
=(n-(n-1)*A1)/(n+1-n*A1)
=(n+1)/(2+n) ,,,,,(n>1)
==>An=n/(n+1)