已知数列{an}的前n项和为Sn,通项an满足Sn+an=1/2(n2+3n-2),求通项公式an.

ai=1/2
Sn=1/2(n2+3n-2)-an
Sn-1=1/2（(n-1)^2+3(n-1)-2）-an-1
相减2an=2n+1+an-1
设参数方程求解后：an-4(n+1)+6=(1/2)^(n-1)(a1-2)
可得an=4n-2+((-3/2)(1/2)^(n-1))