高一数学题1、已知数列an满足a1=4/3,2-a（n+1）=12/(an+6),1/an的前n项和为Sn,求Sn
问题描述：
高一数学题1、已知数列an满足a1=4/3,2-a（n+1）=12/(an+6),1/an的前n项和为Sn,求Sn
2、已知数列an=4n-3,设bn=2/(an·a(n+1)),Tn是数列bn的前n项和,求使得Tn＜m/20对所有n∈N都成立的最小正整数m
答
1、2-a(n+1)=12/(an+6)
a(n+1) = 2an/(an+6)
1/a(n+1) = (an+6)/[2an]
1/a(n+1) + 1/4 = 3(1/an + 1/4)
[1/a(n+1) + 1/4] / (1/an + 1/4) = 3
(1/an + 1/4)/ (1/a1+1/4) = 3^(n-1)
(1/an + 1/4) = 3^(n-1)
1/an = 3^(n-1) -1/4
1/a1+1/a2+..+1/an
= (3^n-1)/2 - n/4
2、bn=2/(an·a(n+1))
=(1/2)*[1/(4n-3)-1/(4n+1)]
Tn=(1/2)*[1-1/5+1/5-1/9+……+1/(4n-3)-1/(4n+1)]
=(1/2)*[1-1/(4n+1)]
=2n/(4n+1)
Tn无限接近于1/2
即m/20>=1/2【因为趋向于0.5即0.5在Tn中不可取所以可以取等】
综上m>=10