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
A3+B5=21
a1+2d+b1q⁴=21
1+2d+q⁴=21即2d+q⁴=20 ①
A5+B3=13
a1+4d+b1q²=13
1+4d+q²=13即4d+q²=12 ②
①*2-②得
2q⁴-q²-28=0
(2q²+7)(q²-4)=0
∵q>0
∴2q²+7>0
∴q²-4=0
∴q=2
代入 ②得 d=2
an=1+2(n-1)=2n-1
bn=2^(n-1)
(2)
an/bn=(2n-1)/2^(n-1)
Sn=1+3/2+5/2²+.+(2n-1)/2^(n-1)
2Sn=2+3+5/2+7/2²+.+(2n-1)/2^(n-2)
相减得
Sn=2+2+2/2+2/2²+...+2/2^(n-2)-(2n-1)/2^(n-1)
=2+2[1+1/2+1/2²+...+1/2^(n-2)]-(2n-1)/2^(n-1)
=2+2[1-1/2^(n-1)]/(1-1/2)-(2n-1)/2^(n-1)
=2+4[1-1/2^(n-1)]-(2n-1)/2^(n-1)
=2+4-4/2^(n-1)-(2n-1)/2^(n-1)
=2+4-4/2^(n-1)-(2n-1)/2^(n-1)
=6-(4+2n-1)/2^(n-1)
=6-(2n+3)/2^(n-1)