数列an满足a1=2an+1=2n+1an/【(n+1/2)an+1+2n】(1)bn=2n/an求bn通向(2)设cn=1/n(n+1)an+1(底n+1项求数列cn的前n项和sn并由此证明5/16≤sn<1/2

来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/23 18:19:30
数列an满足a1=2an+1=2n+1an/【(n+1/2)an+1+2n】(1)bn=2n/an求bn通向(2)设cn=1/n(n+1)an+1(底n+1项求数列cn的前n项和sn并由此证明5/16

数列an满足a1=2an+1=2n+1an/【(n+1/2)an+1+2n】(1)bn=2n/an求bn通向(2)设cn=1/n(n+1)an+1(底n+1项求数列cn的前n项和sn并由此证明5/16≤sn<1/2
数列an满足a1=2an+1=2n+1an/【(n+1/2)an+1+2n】(1)bn=2n/an求bn通向(2)设cn=1/n(n+1)an+1(底n+1项
求数列cn的前n项和sn并由此证明5/16≤sn<1/2

数列an满足a1=2an+1=2n+1an/【(n+1/2)an+1+2n】(1)bn=2n/an求bn通向(2)设cn=1/n(n+1)an+1(底n+1项求数列cn的前n项和sn并由此证明5/16≤sn<1/2
((n+1)^2+1]/n(n+1)=(n²+2n+2)/(n²+n)=1+(n+2)/(n²+n)=1+1/(n-1+2/(n+2))
=1+1/((n+2)+2/(n+2)-3)
(n+2)+2/(n+2)是个对钩函数,最低点是(n+2)=2/(n+2)此时n解出来小于0
所以(n+2)+2/(n+2)是个单调递增函数最小值n=2 所以原式最大值9/2(我令n>=2)
so
1+1/((n+2)+2/(n+2)-3)《5/3(n》2)
n=1时
原式=5/2x1/8=5/16
所以s1=5/16《sn
当n》2时
sn=s1+(c2+.cn)