已知数列{an}的前n项和为Sn,且S(n+1)=4an+2,a1=1,设Cn=an/2^n,求证数列{Cn}是等差数列

S(n+1)=4an+2
Sn=4a(n-1)+2
a(n+1)=S(n+1)-Sn=4(an-a(n-1))
a(n+1)-2an=2[an-2a(n-1)]
[a(n+1)-2an]/[an-2a(n-1)]=2
{an-2a(n-1)}是公比为2的等差数列

a1=1
S2=4+2=6,a2=S2-a1=6-1=5
an-2a(n-1)=[a2-2a1]*2^(n-2)=3*2^(n-2)
an/2^n=12+a(n-1)/2^(n-1)

a(n-1)-2a(n-2)=3*2^(n-3)
a(n-2)/2^(n-2)=a(n-1)/2^(n-1)-12

an/2^n+a(n-2)/2^(n-2)
=(12+a(n-1)/2^(n-1))+(a(n-1)/2^(n-1)-12)
=2a(n-1)/2^(n-1)
数列{Cn}是等差数列