作业帮已知函数f(x)=1/2+log2x/(1-x）,若Sn=f(1/n)+f(2/n)+.+f((n-1)/n),n为正整数,且n≥2,（1）求Sn.（2）a1=2/3,an=1/[(Sn+1)（Sn下一项+1）]（n≥2,n∈正整数）数列an的前n项和为Tn,若Tn＜λ（Sn的下一项+1）对一切n∈N*
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![已知函数f(x)=1/2+log2x/(1-x）,若Sn=f(1/n)+f(2/n)+.+f((n-1)/n),n为正整数,且n≥2,（1）求Sn.（2）a1=2/3,an=1/[(Sn+1)（Sn下一项+1）]（n≥2,n∈正整数）数列an的前n项和为Tn,若Tn＜λ（Sn的下一项+1）对一切n∈N*](/uploads/image/z/4948284-12-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D1%2F2%2Blog2x%2F%281-x%EF%BC%89%2C%E8%8B%A5Sn%3Df%281%2Fn%29%2Bf%282%2Fn%29%2B.%2Bf%28%28n-1%29%2Fn%29%2Cn%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E4%B8%94n%E2%89%A52%2C%EF%BC%881%EF%BC%89%E6%B1%82Sn.%EF%BC%882%EF%BC%89a1%3D2%2F3%2Can%3D1%2F%5B%28Sn%2B1%29%EF%BC%88Sn%E4%B8%8B%E4%B8%80%E9%A1%B9%2B1%EF%BC%89%5D%EF%BC%88n%E2%89%A52%2Cn%E2%88%88%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E8%8B%A5Tn%EF%BC%9C%CE%BB%EF%BC%88Sn%E7%9A%84%E4%B8%8B%E4%B8%80%E9%A1%B9%2B1%EF%BC%89%E5%AF%B9%E4%B8%80%E5%88%87n%E2%88%88N%2A)
已知函数f(x)=1/2+log2x/(1-x）,若Sn=f(1/n)+f(2/n)+.+f((n-1)/n),n为正整数,且n≥2,（1）求Sn.（2）a1=2/3,an=1/[(Sn+1)（Sn下一项+1）]（n≥2,n∈正整数）数列an的前n项和为Tn,若Tn＜λ（Sn的下一项+1）对一切n∈N*
已知函数f(x)=1/2+log2x/(1-x）,若Sn=f(1/n)+f(2/n)+.+f((n-1)/n),n为正整数,且n≥2,（1）求Sn.
（2）a1=2/3,an=1/[(Sn+1)（Sn下一项+1）]（n≥2,n∈正整数）数列an的前n项和为Tn,若Tn＜λ（Sn的下一项+1）对一切n∈N*都成立,求λ的取值范围
已知函数f(x)=1/2+log2x/(1-x）,若Sn=f(1/n)+f(2/n)+.+f((n-1)/n),n为正整数,且n≥2,（1）求Sn.（2）a1=2/3,an=1/[(Sn+1)（Sn下一项+1）]（n≥2,n∈正整数）数列an的前n项和为Tn,若Tn＜λ（Sn的下一项+1）对一切n∈N*
1、
f(k/n)=1/2+log2[(k/n)/(1- k/n)]=1/2+log2[k/(n-k)]
∴f(1/n)=1/2+log2[1/(n-1)],f(2/n)=1/2+log2[2/(n-2)],f(3/n)=1/2+log2[3/(n-3)]……
Sn=1/2+log2[1/(n-1)]+1/2+log2[2/(n-2)]+1/2+log2[3/(n-3)]+……+1/2+log2{(n-1)/[n-(n-1)]}
=(n-1)/2+log2[1/(n-1)]+log2[2/(n-2)]+log2[3/(n-3)]+……+log2{(n-1)/[1]}
=(n-1)/2+log2{[1/(n-1)]*[2/(n-2)]*[3/(n-3)]……*[(n-1)/1]}
=(n-1)/2+log2(1)
=(n-1)/2
2、
an=1/[(Sn+1)·(S(n+1)+1)]
Sn+1=(n+1)/2 S(n+1)+1=n/2+1=(n+2)/2
an=4/(n+1)(n+2)
Tn
= 2/3 + 4/(3*4) + 4/(4*5) + …… + 4/(n+1)(n+2)
=4/(2*3) + 4/(3*4) + 4/(4*5) + …… + 4/(n+1)(n+2)
=4[1/(2*3)+1/(3*4)+1/(4*5)+……+1/(n+1)(n+2)]
=4[1/2 - 1/2 + 1/3-1/4+1/4-1/5……+1/(n+1)-1/(n+2)]
=4[1/2-1/(n+2)]
=2n/(n+2)
＜λ(S(n+1)+1)
=λ(n+2)/2
2n/(n+2)＜λ(n+2)/2 整理得
(n+2)²/n＞4/λ
∵(n+2)²/n=(n²+4n+4)/n = n + 4/n + 4 ≥2√(n * 4/n) + 4 =8 当且仅当n=2时,等号成立.
∴4/λ＜8
∴λ＞1/2