1.
bn=1/(an-1/2)
an=1/bn +1/2
代入
8a(n+1)an-16a(n+1)+2an+5=0
8/bnbn+1) +2+4/bn+4/bn+1) -16/bn+1) -8+2/bn +1+5=0
8+4bn+1)+4bn-16bn+2bn+1)=0
8-12bn+6bn+1)=0
6bn+1)=12bn-8
bn+1)=2bn/ -4/3
设bn+1)+k=2(bn+k)
解得：k=-4/3
bn+1)-4/3=2(bn-4/3)
bn+1)-4/3)/(bn-4/3)=2
2为常数
所以bn-4/3为等比数列.
b1=1/(a1-1/2)=2
bn-4/3=(2-4/3)*2^(n-1)
bn=(2^n)/3+4/3
an=1/bn +1/2
anbn=1+ bn/2=2^(n-1) /3 +5/3
S(anbn)=(1/3)*(1-2^n)/(1-2) +5n/3
=(2^n)/3 +5n/3 -1/3