a（n+1）=(-1/2)an+ n 求{an}

n=(2/3)(n+1)+(1/3)n-2/3;
a(n+1)=(-1/2)an+n=(-1/2)an+(2/3)(n+1)+1/3n-2/3;
a(n+1)-(2/3)(n+1)=(-1/2)[an-(2/3)n]-2/3;
令bn=an-(2/3)n;
则：
b(n+1)=(-1/2)bn-2/3;
bn(n+1)+4/9=(-1/2)(bn+4/9);
令cn=bn+4/9;
则：cn为公比（-1/2)的等比数列；
cn=c1*(-1/2)^(n-1);
cn=bn+4/9;c1=b1+4/9;
所以：
bn+4/9=(b1+4/9)(-1/2)^(n-1)；
bn=an-(2/3)n;
b1=a1-2/3;
an-(2/3)n+4/9=(a1-2/3+4/9)(-1/2)^(n-1);
an=(a1-2/9)(-1/2)^(n-1)+(2/3)n-4/9;
(首项a1题设未给出，故为最终结果）；