已知数列｛an｝,Sn是它的前n项和,并且S(n+1)=4an+2 ,a1=1,

S(n+1)=4an+2
Sn=4a(n-1)+2
两式相减得
a(n+1)=4an-4a(n-1)
a(n+1)-2an=2an-4a(n-1)
a(n+1)-2an=2[an-2a(n-1)]
[a(n+1)-2an]/[an-2a(n-1)]=2
因为bn=a(n+1)-2an
所以bn/b(n-1)=2
所以{bn}是以2为公比的等比数列
S(n+1)=4an+2
S2=a1+a2=4a1+2
a2=3a1+2
a2=3*1+2=5
b1=a2-2a1
b1=5-2*1=3
bn=b1q^(n-1)
=3*2^(n-1)
cn=an/2^n
c(n+1)=a(n+1)/2^(n+1)
c(n+1)-cn=a(n+1)/2^(n+1)-an/2^n
=a(n+1)/2^n-2an/2^(n+1)
=[a(n+1)-2an]/2^(n+1)
=bn/2^(n+1)
=3*2^(n-1)/2^(n+1)
=3*2^(n-1-n-1)
=3*2^(-2)
=3/4
所以{cn}是以3/4为公差的等差数列

（1）由S(n+1) = 4an+2 可得 Sn = 4a(n-1)+2 (n>=2)两式相减可得 a(n+1)=4an-4a(n-1)
化简得 a(n+1)-2an = 2(an-a(n-1))
故 bn = 2b(n-1) (n>=2) bn为等比数列
又S2=4a...