数列Cn=[(n+1)^2+1]/[n(n+1)*2^(n+2)],Sn是其前n项和,求证:5/16
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![数列Cn=[(n+1)^2+1]/[n(n+1)*2^(n+2)],Sn是其前n项和,求证:5/16](/uploads/image/z/3955650-42-0.jpg?t=%E6%95%B0%E5%88%97Cn%3D%5B%28n%2B1%29%5E2%2B1%5D%2F%5Bn%28n%2B1%29%2A2%5E%28n%2B2%29%5D%2CSn%E6%98%AF%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E6%B1%82%E8%AF%81%EF%BC%9A5%2F16)
数列Cn=[(n+1)^2+1]/[n(n+1)*2^(n+2)],Sn是其前n项和,求证:5/16
数列Cn=[(n+1)^2+1]/[n(n+1)*2^(n+2)],Sn是其前n项和,求证:5/16<=Sn<1/2
数列Cn=[(n+1)^2+1]/[n(n+1)*2^(n+2)],Sn是其前n项和,求证:5/16
((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)
5/16+5/3x1/8=25/48>1/2!我也尝试过这种方法,都试到c4去了,就是想放大得小点,结果还是大于1/2