已知数列an的前n项和Sn=n-5an-85(1)求an的通项公式（2）令bn=log5/6×1-a1/18+log5/

Sn=n-5an-85 (1)
S（n+1）=n+1-5a（n+1）-85 (2)
(2)-(1)整理得6a(n+1)=1+5an
即a(n+1)-1=(5/6)(an-1)
又由S1=a1=1-5a1-85得a1=-14
所以{an-1}为首项-15,公比5/6的等比数列
所以an=(-15)*（5/6）^(n-1)+1
(2)bn=log5/6(5/6*(5/6)^0)+log5/6(5/6*(5/6)^1)+log5/6(5/6*(5/6)^2)+...+log5/6(5/6*(5/6)^(n-1))
=1+0+1+1+1+2+1+3+...+1+(n-1)
=n+(0+n-1)*n/2
=n(2+n-1)/2
=n(n+1)/2
1/bn=2/(n(n+1))=2[1/n-1/(n+1)]
Tn=2[1-1/2+1/2-1/3+,.+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)

n=1时，a1=S1=1-5a1-85
解得 a1=-14；
n≥2时，
an=Sn-S(n-1)
=n-5an-85-(n-1)+5a(n-1)+85
=1-5an+5a(n-1)
则 6an=1+5a(n-1)
设存在x∈R使 6[a(n)-x]=5[a(n-1)-x]
则 x=1，an-1=(5/6)[a(n-1)-1]