已知等差数列{an}的首项a1=1,公差d=1,前n项和为Sn,bn=1/Sn, (1)求数列{bn}的通项公式; (2)求证:b1+b2+…+bn<2.

问题描述:

已知等差数列{an}的首项a1=1,公差d=1,前n项和为Snbn

1
Sn

(1)求数列{bn}的通项公式;
(2)求证:b1+b2+…+bn<2.

(1)∵等差数列{an}中a1=1,公差d=1∴Sn=na1+n(n−1)2d=n2+n2∴bn=2n2+n…(4分)(2)∵bn=2n2+n=2n(n+1)…(6分)∴b1+b2+b3+…+bn=2(11×2+12×3+13×4+…+1n(n+1))=2(1−12+12−13+13−14+…+1n−1n+1)…...