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
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![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](/uploads/image/z/2718688-40-8.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D2an%2B1%3D2n%2B1an%2F%E3%80%90%28n%2B1%2F2%29an%2B1%2B2n%E3%80%91%EF%BC%881%EF%BC%89bn%3D2n%2Fan%E6%B1%82bn%E9%80%9A%E5%90%91%EF%BC%882%EF%BC%89%E8%AE%BEcn%3D1%2Fn%EF%BC%88n%2B1%EF%BC%89an%2B1%EF%BC%88%E5%BA%95n%2B1%E9%A1%B9%E6%B1%82%E6%95%B0%E5%88%97cn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Csn%E5%B9%B6%E7%94%B1%E6%AD%A4%E8%AF%81%E6%98%8E5%2F16%E2%89%A4sn%EF%BC%9C1%2F2)
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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)
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