设数列an的前n项和为sn,a1=1/4,且2an=2an-1+1,数列bn,满足b1=3/4,且3bn-bn-1=n求证bn-an为等比数列和bn的通项公式

Question

设数列an的前n项和为sn,a1=1/4,且2an=2an-1+1,数列bn,满足b1=3/4,且3bn-bn-1=n 求证bn-an为等比数列和bn的通项公式

Answer 1

2an=2a(n-1) +1
an = a(n-1) +1/2
an -a(n-1) =1/2
{an}是等差数列,d=1/2
an -a1 = (n-1)/2
an = (2n-1)/4
3bn-b(n-1)=n
bn -(1/2)n + 1/4= (1/3)[ b(n-1) -(1/2)(n-1) + 1/4]
{bn -(1/2)n + 1/4} 是等比数列,q= 1/3
bn -(1/2)n + 1/4 = (1/3)^(n-1).( b1 - 1/2 +1/4)
= (1/2).(1/3)^(n-1)
bn = (1/2)n -1/4 +(1/2).(1/3)^(n-1)
bn -an =(1/2).(1/3)^(n-1)
=>{bn-an}是等比数列