设{an}是等差数列，{bn}是各项都为正数的等比数列且a1=b1=1,a3+b5=21,a5+b3=13

(1)设公差为d，公比为q，显然q>0
则2d+q^4=20 (1) 4d+q^2=12 (2)
(1)*2-(2) (2q^2+7)(q^2-4)=0
∵q>0
∴q=2 代入得d=2
an=1+2(n-1)=2n-1
bn=2^(n-1)
(2)

an/bn=(2n-1)/2^(n-1)=(4n-2)/(2^n)
错项相消法
Sn=a1/b1+a2/b2+a3/b3+……+an/bn ......................①
=2/2+6/2²+10/2³+……+(4n-2)/(2^n)
1/2Sn=2/2²+6/2³+……+4/(2^n)-【(4n-2)/(2^n+1)】 ......................②
①-②得：
1/2Sn=1+【4/2²+4/2³+……+4/(2^n)】-【(4n-2)/(2^n+1)】

=1+【4/2²-4/(2^n)*1/2】/(1-1/2)-【(4n-2)/(2^n+1)】

=1+【2-4/(2^n)】-【(4n-2)/(2^n+1)】
=1+2-4/(2^n)-【(2n-1)/(2^n)】

Sn=6-(4n+6)/(2^n)

• 易求q=2，d=2,an=2n-1,bn=2的n-1次方