数列{an}中,a1=2/3,若函数y=3x-1过点(an+1,an)求证:数列an-1/2为等比数列 2求数列

∵函数y=3x-1过点(an+1,an)
∴an=3a(n+1)-1
an-1/2=3[a(n+1)-1/2]
[a(n+1)-1/2]/[an-1/2]=1/3
所以数列an-1/2是以首项a1-1/2=1/6,公比q=1/3的等比数列
2、
令bn=an-1/2
则b(n+1)/bn=1/3,b1=a1-1/2=1/6
所以bn=1/6×(1/3)^(n-1)=1/2×1/3^n
即an-1/2=1/6×(1/3)^(n-1)=1/2×1/3^n
∴an=1/2×1/3^n+1/2sn怎么求】Sn=a1+a2+a3+……+an=(1/2×1/3+1/2)+(1/2×1/3²+1/2)+(1/2×1/3³+1/2)+……+(1/2×1/3^n+1/2)=1/2(1/3+1/3²+1/3³+1/3^n)+n/2 =1/2×1/3[1-(1/3)^n]/(1-1/3)+n/2 =1/4×[1-(1/3)^n]+n/2 不懂再追问