Question

已知等比数列｛an｝的各项均为正数，且2a1+3a2=1，a3 +9a2a6.

(1)求an的通项公式 (2）设bn=log3a1+log3a2+……+log3an,求{1/bn}的前n项和 【请帮我完成这道题，并写出解答过程和分析！】

Answer 1

(1)a3^2=9a2a6 (a2p)^2=9a2(a2p^4) a2^2p^2=9a2^2p^4 ∵此数列各项均为正数∴a2^2<>0,p>0 两边同时除以a2^2p^2，得9p^2=1,p=1/3 2a1+3a2=1 2a1+3*[(1/3)*a1]=1 2a1+a1=1 3a1=1 a1=1/3 an=a1p^(n-1)=1/3*(1/3)^(n-1)=1/3^n (2)bn=log3a1+log3a2+...+log3an =log3(a1*a2*...*an) =log3[(1/3)*(1/3^2)*...*(1/3^n)] =log3[(1/3)^(1+2+...+n)] =(1+2+...+n)*log3(1/3) =-n(n+1)/2 1/bn=-2/n(n+1) =(-2)*[1/n(n+1)] =(-2)*[1/n-1/(n+1)] 1/b1+1/b2+...+1/bn =(-2)*(1-1/2)+(-2)*(1/2-1/3)+...+(-2)*[1/n-1/(n+1)] =(-2)*[1-1/2+1/2-1/3+...+1/n-1/(n+1)] =(-2)*[1-1/(n+1)] =-2n/(n+1)

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