设数列{an}是等差数列,{bn}为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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设数列{an}是等差数列,{bn}为各项都为正数的等比数列.且a1=b1=1,a3+b5=21,a5+b3=13.
（1）求数列{an}、{bn}的通项公式.
（2）求数列{an/bn}的前几项和.

an=(n-1)d+1
bn=q^(n-1)
2d+1+q^4=21
4d+1+q^2=13
2*q^4-q^2=28
(2q^2+7)(q^2-4)=0
q^2=4
因为q大于零,所以
q=2,d=2
an=2n-1
bn=2^(n-1)
第二题有点麻烦
一个等差数列乘以一个等比数列有个通用的方法
列出Sn作为一式,Sn*q作为二式
二式减一式即可推导出Sn,你自己算下吧

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