等比数列{an}中,首项a1>0,公比q>0,其前n项和为Sn,求证lgSn+lgSn+2

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等比数列{an}中,首项a1>0,公比q>0,其前n项和为Sn,求证lgSn+lgSn+2

等比数列{an}中,首项a1>0,公比q>0,其前n项和为Sn,求证lgSn+lgSn+2
等比数列{an}中,首项a1>0,公比q>0,其前n项和为Sn,求证lgSn+lgSn+2

等比数列{an}中,首项a1>0,公比q>0,其前n项和为Sn,求证lgSn+lgSn+2
Sn=a1(1-q^n)/(1-q)
S(n+1)=a1[1-q^(n+1)]/(1-q)
S(n+2)=a1[1-q^(n+2)]/(1-q)
Sn*S(n+2)=a1^2*(1-q^n)[1-q^(n+2)]/(1-q)^2
=[a1^2/(1-q)]^2*[1-q^n-q^(n+2)+q^(2n+2)] (1)
S(n+1)*S(n+1)=[a1^2/(1-q)]^2*[1-2q^(n+1)+q^(2n+2)] (2)
∵(1-q)^2>0(通常q≠1)
∴1+q^2>2q
∴q^n+q^(n+2)>2q^(n+1)
对比(1)(2)式知 Sn*S(n+2)