设数列{an}是等差数列数列,{bn}是各项都为正数的等比数列,且a1=b1=1,b1+b2=a2,b3是a1与a4的等

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设数列{an}是等差数列数列,{bn}是各项都为正数的等比数列,且a1=b1=1,b1+b2=a2,b3是a1与a4的等差中项
（1）求数列{an}、{bn}的通项公式（2）求数列{an/bn}的前n项和

∵{an}是等差数列数列,{bn}是各项都为正数的等比数列
∴b1+b2=a2
∵a1=b1=1
即1+q=1+d ∴q=d
又∵2b3=a1+a4 ∴2q²=2+3d
结合q=d得
q=2或q=-1/2
∵bn各项均为正数
∴q=-1/2(舍去) q=d=2
∴an=1+2(n-1)=2n-1
bn=2^(n-1)
令Tn=an/bn=(2n-1)/(2^(n-1))
应该是会用到错位相减法,太麻烦了,不算了.

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