S(n+1)＝4an+2;Sn＝4a(n-1)+2
a(n+1)=S(n+1)-Sn＝4an-4a(n-1)
a(n+1)=4[an-4a(n-1)]
即：a(n+1)-2an＝2[an-2a(n-1)]
[a(n+1)-2an]/[an-2a(n-1)]＝2
bn/b(n-1)＝2
{bn}为等比数列,q=2,
S2=4a1+2=6=a1+a2,a2=5,b1=a2-2a1=3
bn=b1*2^(n-1)=3*2^(n-1)
a(n+1)-2an=3*2^(n-1)
a(n+1)＝2an+3*2^(n-1)
an的通项要用归纳法证明!