a1=2,an=(3-a(n-1))/2,求an

an=-a(n-1)/2+(3/2)
an+x=-a(n-1)/2+(3/2)+x=-1/2[a(n-1)-3-2x]
令x=-3-2x
x=-1
所以an-1=-1/2[a(n-1)-1]
(an-1)/[a(n-1)-1]=-1/2
所以an-1是等比数列,q=-1/2
a1=2,a1-1=1
所以an-1=(a1-1)*q^(n-1)=(-1/2)^(n-1)
an=1+(-1/2)^(n-1)

2an=3-a(n-1)
2(an+t)=-[a(n-1)+t]
2an+2t=-a(n-1)-t
t=-1
2(an-1)=-[a(n-1)-1]
令an-1=bn
2bn=-b(n-1)
bn/b(n-1)=-1/2
{bn}是公比为-1/2的等比数列
bn=b1*(-1/2)^(n-1)
bn=(-1/2)^n
an=(-1/2)^n+1