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
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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
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
![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](/uploads/image/z/2559206-38-6.jpg?t=%E9%AB%98%E4%B8%80%E6%95%B0%E5%AD%A6%E9%A2%981%E3%80%81%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D4%2F3%2C2-a%EF%BC%88n%2B1%EF%BC%89%3D12%2F%28an%2B6%29%2C1%2Fan%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%B1%82Sn)
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
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
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