1,已知各项均为正数的等差数列{an}的首项a1=1,前n项和为Sn,且满足关系ana（n+1）=4Sn-1,

1
设{an}公差为d
则an=1+(n-1)d
Sn=n+n(n-1)d/2
∵ana（n+1）=4Sn-1
∴[1+(n-1)d][1+nd]=4n+2n(n-1)d-1
∴1+(2n-1)d+n(n-1)d²=(4-2d)n+2dn²-1
∴d²n²+(2d-d²)n+1-d=(4-2d)n+2dn²-1
∴左右恒等,同类项系数相等
∴常数项1-d=-1,d=2此时n,n²系数也相等
∴d=2
an=2n-1
2
bn=1/{（an+1)[a（n+1)+1]}
=1/[2n(2n+2)]
=1/4*1/[n(n+1)]
=[1/n-1/(n+1)]*1/4
∴Tn=1/4*[1-1/2+1/2-1/3+1/3-1/4+.+1/n-1/(n+1)]
=1/4[1-1/(n+1)]
=n/(4n+4)