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.

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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.
（一）求｛an｝、｛bn｝的通项公式；（二）求数列｛an／bn｝的前n项和Sn.

设
an=a1+(n-1)d
bn=b1(n-1)^q
a1=b1=1.(1)
a5+b3=13.(2)
a3+b5=21.(3)
4d+q^2=12.(4)
2d+q^4=20.(5)
(5)*2-(4)得
2q^4-q^2-28=0
(q^2-4)(2q^2+7)=0
q^2=4
q=2
d=2
an=2n-1
bn=2^(n-1)
Sn=1/1+3/2+5/4+...+(2n-1)/2^(n-1)
(1/2)S=1/2+3/4+5/8+...+(2n-1)/2^n
相减得
(1/2)S=1+1+1/2+...+1/2^(n-2)-(2n-1)/2^n
=3-1/2^(n-2)-(2n-1)/2^n
=3-(2n+3)/2^n
Sn=6-(2n+3)/2^(n-1)
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