(1)bn=log3(an-n),a4=31,b4=log3(a4-4)=log3(31-4)=3,由{bn}是递增的等差数列,能求出数列{bn}的通项公式.
(2)bn=n-1=log3(an-n),所以an=3n-1+n,由此能求出数列{an}的前n项和.
(1)bn=log3(an-n),a4=31,
b4=log3(a4-4)=log3(31-4)=3,
∵{bn}是递增的等差数列
∴b3+b5=2b4=6,
b3b5=8,
解得b3=2,b5=4,
d=1,b1=0,
数列{bn}的通项公式:bn=n-1.
(2)bn=n-1=log3(an-n)
所以an=3n-1+n
Sn=3+31+32+…+3n-1+1+2+3+…+n
=.