题目

数列题,求证明
数列{an}中,a1=1/4,a2=3/4,2an=an+1 +an-1 数列{bn}中,b1小于0,3bn-bn-1=n 前n项和为Sn
求证{bn-an}是等比数列
求证{bn}是递增数列

提问时间：2021-04-02

答案

2an=an+1 +an-1 -> {an}是等差数列
a1=1/4,a2=3/4 -> an = (2n - 1) / 4
3an - an-1 = (6n - 3) / 4 - (2n - 3) / 4 = n = 3bn-bn-1
-> 3(bn - an) = bn-1 - an-1
-> {bn-an}是等比数列
∵ b1 < 0 ,a1 > 0
∴ b1 - a1 < 0
∴ bn - an = (b1 - a1) * (1/3)^(n-1) < 0
∴ bn - an < 3 * (bn - an) = bn+1 - an+1 = bn+1 - (an + 1/2)
∴ bn < bn+1 - 1/2 < bn+1
∴ {bn}是递增数列