设{an}是等差数列 {bn}是各项都为正数的等比数列,且a1=b1=1,a2+b3=a3+b2=7 (1)求{an},

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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,a2+b3=a3+b2=7 (1)求{an},{bn}的通项公式 ...
设{an}是等差数列 {bn}是各项都为正数的等比数列,且a1=b1=1,a2+b3=a3+b2=7 (1)求{an},{bn}的通项公式 (2)求数列{an/bn}的前n项和Sn 急

公差d,公比q代入得：d=q=2
an=2n-1;bn=2^(n-1)
an/bn=(2n-1)/2^(n-1)
Sn=1+3/2+5/2^2+……+(2n-1)/2^(n-1)
Sn/2=1/2+3/2^2+5/2^3+……+(2n-3)/2^(n-1)+(2n-1)/2^n
二式相减,右侧错位:
Sn=6- (2n+3)/2^(n-1)

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