An=2A（n-1）＋ 3*2^n
A2=2A1+3*2^2=2+3*2^2
A3=2A2+3*2^3=2^2+6*2^3
A4=2A3+3*2^4=2^3+9*2^4
A5=2A4+3*2^5=2^4+12*2^5
……………………………
猜测：
An=2^(n-1)+3(n-1)*2^n
归纳法证明!
(1)n=1时,显然成立!
(2)假设n=k时,Ak=2^(k-1)+3(k-1)*2^k
n=k+1时,
A(k+1)=2Ak+3*2^(k+1)=2^k+6(k-1)*2^k+3*2^(k+1)=2^k+3k*2^(k+1)
也成立!
综上,An=2^(n-1)+3(n-1)*2^n＝(6n-5)*2^(n-1)