在数列{An}中，Sn是它的前n项和，A1=1 且
S<n+1>=4An+2
(1)设Bn=A<n+1>-2An,求证数列{Bn}是等比数列
(2)设Cn=An/2^n，求证数列{An}是等差数列
(3)求数列{An}的通项公式及前n项和公式
请把过程写的详细一些，谢谢了
“<>”中的内容为右下角标

1. S(n+1)=4*An +2 ...(1), Sn =4*A(n-1) +2 ...(2)
(1)-(2): A(n+1) =S(n+1)-Sn =4*An -4*A(n-1)
==> [A(n+1) -2An]/[An -2*A(n-1)] =2
==> Bn =A(n+1)-2An是等比数列，公比q=2
A2 =4*A1 +2 =6, B1 =A2-2*A1 =4
2. A(n+1) -2An =Bn =4*2^(n-1) =2^(n+1)
==> A(n+1)/2^(n+1) - An/2^n =1
==> Cn=An/2^n是等差数列。（题目有误）
{Cn}的公差=1，首项C1=1/2
3. Cn =C1+(n-1)*1 = n -1/2
==> An =Cn*2^n =n*2^n -2^(n-1)
==> Sn =4*A(n-1) +2 =(n-1)*2^(n+1) -2^n +2