数列2a(n+1)-an=n a1=1/2 求数列an的通项公式

2an+1=an + n
2an+2=an+1 + n+1
2an+2-2an+1=an+1-an+1 令bn=an+1-an
2bn+1=bn + 1 bn+1 -1 = 1/2 (bn -1),bn-1为首项a2-a1-1,公比1/2的等比
bn=1-3*(1/2)^(n+1)
an-an-1=1-3*(1/2)^n
...
a2-a1=1-3*1/4
叠加,an -1/2 =n-1+3*(1/4+.(1/2)^n)
an=n-2+3*(1/2)^n (n>=2)
经验证,a1也满足,所以通项同上

an=-(1/2)^n+1