已知数列{an}是首项a1=1的等比数列且an>0{bn}是首项为1的等差数列又a5+b3=21,a3+b5=13

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已知数列{an}是首项a1=1的等比数列且an>0{bn}是首项为1的等差数列又a5+b3=21,a3+b5=13
求数列an和bn的通项公式

an=q^(n-1),bn=1+(n-1)d
可得q^4+1+2d=21,q^2+1+4d=13
解得q^2=4,d=2
又q>0故q=2
所以an=2^(n-1),bn=2n-1

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