设等差数列{an}的前n项和为Sn,且满足条件S8=36,a3=3.
(1)求数列{an}的通项公式;
(2)令bn=[1an
(1)∵等差数列{an}的前n项和为Sn,且满足条件S8=36,a3=3,
8a1+
2d=36
a1+2d=3,
解得a1=1,d=1,∴an=n.
(2)∵bn=[1
an+
an+1+…+
a2n=
1/n+
n+1+…+
2n],
∴bn+1-bn=[1/2n+2+
2n+1−
n]
=([1/2n+2−
2n])+([1/2n+1−
2n])