已知数列{an}满足a1=1/4,a2=3/4,a(n+1)=2an-a(n-1)(n>等于2,n属于N*),数列{bn

（1）
a(n+1)-an=an-a(n-1)
数列{an}为等差数列
d=a2-a1=1/2
an=a1+(n-1)/2=n/2-1/4
(2)
3b2-b1=2,b2=2/3+b1/3
3b3-b2=3,b3=3+2/3+b1/3=11/3+b1/3
3b4-b3=4,b4=4+11/3+b1/3=23/3+b1/3
3b5-b4=5,b5=5+23/3+b1/3=38/3+b1/3
.
bn=b1/3+[-7/3+3*(1+2+...+n)/3]
=b1/3-7/3+(1+n)n/2
=n(n+1)/2+(2b1-14)/6
n>=2
bn-an=n^2/2+n/2-n/2+1/2-7/3+b1/3
=n^2/2+(2b1-11)/6
b(n+1)-a（n+1)=b1/3-7/3+(n+1)(n+2)/2-(n+1)/2+1/2
=(n+1)^2/2+(2b1-11)/6
要成等比,必须b1=11/2
(3).