已知数列{an}满足,anan+1=n(n-1)(an+1-an),且a1=0,a2=1. (1)求证:数列{an}是等差数列; (2)设bn=2 an-34,求数列{|bn|}的前n项和Sn.

问题描述:

已知数列{an}满足,anan+1=n(n-1)(an+1-an),且a1=0,a2=1.
(1)求证:数列{an}是等差数列;
(2)设bn=2 an-34,求数列{|bn|}的前n项和Sn

(1)证明:∵a1=0,a2=1,依题意只需证明∀n∈N*,an+1-an=a2-a1=1,…(1分)∵anan+1=n(n-1)(an+1-an),∴an+1=n(n−1)ann(n−1)−an,n>1,∴只需证an+1−an=an2n(n−1)−an=1,n>1.…(3分)即只需证...