已知数列{An}前n项和为Sn,A1=10,A（n+1）=9Sn+10.1,求证{lgAn}是等差数列.

An=9Sn-1+10,An+1-An=9Sn-9Sn-1=9An,所以,An=10An-1,A1=10,所以,An=10^n,lgAn=n.
2问,设Cn=3/（lgAn)(lgA(n+1),则Cn=3/n（n+1),Tn=3（1/1*2+1/2*3+1/3*4+.+1/n(n+1)）,Tn=3（2-1/1*2+3-2/2*3+4-3/3*4+.+n+1-n/n(n+1)）=3（1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/n - 1/(n+1)）=3（1- 1/(n+1)）=3n / (n+1)