在数列an中,a1=1/2,2a(n+1)=an+n

反正最后都要求An,所以先把An求出来
∵2A(n+1)=An+n
∴A(n+1)=(1/2)(An+n)①
设x,y为常数（所以x、y后面的（n+1）不是下标哦）
令A(n+1)+x(n+1)+y=(1/2)(An+xn+y) ②
A(n+1)=(1/2)(An+xn+y)-x(n+1)-y
=(1/2)An+(-1/2)xn+(-1/2)y-x
与①对照系数,有
(-1/2)xn=(1/2)n
(-1/2)y-x=0
求得x=-1,y=2
把结果代入②,有：
A(n+1)-(n+1)+2=(1/2)(An-n+2)
即[A(n+1)-(n+1)+2]/(An-n+2)=1/2
数列An-n+2是等比数列,公比为1/2,首项A1-1+2=3/2
则An-n+2=(3/2)*[(1/2)^(n-1)]
则An=n-2+(3/2)*[(1/2)^(n-1)]
(1)Bn=A(n+1)-An-1
={n-1+(3/2)*[(1/2)^n]}-{(n-2)+(3/2)*[(1/2)^(n-1)]}-1
=(3/2)(1/2^n)-[1/2^(n-1)]
=(-3/4)/[2^(n-1)]
则Bn是以-3/4为首项,1/2为公比的等比数列
(2)An=n-2+(3/2)*[(1/2)^(n-1)]