数列{an}和{bn}满足a1=1 a2=2 an>0 bn=根号an*an+1

学习时间:2026-04-02 19:07:37 阅读：8955

数列{an}和{bn}满足a1=1 a2=2 an>0 bn=根号an*an+1且{bn}是以公比为q的等比数列(1)证明an+2=anq^2 (2)Cn=a2n-1+2a2n证明{Cn}是等比数列

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2026-04-02 19:07:37

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n=√an*a(n+1) b(n+1)=√a(n+1)a(n+2)[b(n+1)/bn]^2=[a(n+1)*a(n+2)]/[a(n+1)*an]=a(n+2)/ana(n+2)=q^2*an