n+1-bn=3*2^(2n-1).
bn-bn-1=3*2^(2n-3) (1)
bn-1-bn-2=3*2^(2n-5) (2)
...
b2-b1=3*2^(2-1) (n-1)
上述（n-1)上式子相加得：
bn=b1+3*(2+4+...+2^(2n-3))=2+3*2*2^(n-1)=2+3*2^n
a3=1+2d
a7=1+6d
a9=1+8d
(1+6d+2)^2=(1+2d)(1+8d)
9+36d+36d^2=1+10d+16d^2
20d^2+26d+8=0
10d^2+13d+4=0
(2d+1)(5d+4)=0
d=-1/2或d=-4/5
与d>0矛盾