数列题,求证明
问题描述：
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
答
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
∴ b1 - a1 ∴ bn - an = (b1 - a1) * (1/3)^(n-1) ∴ bn - an ∴ bn ∴ {bn}是递增数列