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Sn2、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
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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Sn2、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](/uploads/image/z/2719107-27-7.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%82Sn2%E3%80%81%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%3D4n-3%2C%E8%AE%BEbn%3D2%2F%28an%C2%B7a%28n%2B1%29%29%2CTn%E6%98%AF%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E6%B1%82%E4%BD%BF%E5%BE%97Tn%EF%BC%9Cm%2F20%E5%AF%B9%E6%89%80%E6%9C%89n%E2%88%88N%E9%83%BD%E6%88%90%E7%AB%8B%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E6%95%B4%E6%95%B0m)
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Sn2、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
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Sn2、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
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
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题详情页