已知数列{An}中,Sn是它的前n项和,并且Sn+1(n+1为下标)=4An+2(n=1,2,3...),A1=1,求a

Sn+1=4an+2,①Sn=4an-1+2,② ①-②得,an+1=4an-4an-1,
an+1-2an=2(an-2an-1),令bn=an-2an-1（an-1表示第n-1项）,则数列{bn}是以a2-2a1=3为首项,以2为公比的等比数列.bn=3×2^n-2(2^n-2表示2的n-2次方）,即an-2an-1=3×2^n-2
两边同时除以2^n,得an/2^n -an-1/2^n-1 =3/4,∴数列{an/2^n}是以a1/2=1/2为首项,以3/4为公差的等差数列.an/2^n=(3n-1)/4,
an=(3n-1)2^n-2,Sn=(3n-4)2^n-1+2