大神,请帮忙,急啊!!!
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证明:题意得
a1>0, q>0且q≠1
Sn*Sn+2=(a1q^n-a1)/(q-1)*[a1q^(n+2)-a1]/(q-1)=a1²/(q-1)²*(q^n-1)(q²*q^n-1)
S²n+1=[(a1q*q^n-a1)/(q-1)]²=a1²/(q-1)²*(q*q^n-1)²
(q^n-1)(q²*q^n-1)-(q*q^n-1)²=[q^(2n+2)-q^n-q^(n+2)+1]-[q^(2n+2)-2q^(n+1)+1]
=q^(2n+2)-q^n-q^(n+2)+1-q^(2n+2)+2q^(n+1)-1
=q^n*(2q-1-q²)
=-q^n*(q-1)²<0
∴(q^n-1)(q²*q^n-1)<(q*q^n-1)²
∴Sn*Sn+2<S²n+1
∵0<a<1
∴loga Sn*Sn+2>loga S²n+1
logaSn+logSn+2>2logaSn+1
∴1/2(logaSn+logaSn+2)>logaSn+2
a1>0, q>0且q≠1
Sn*Sn+2=(a1q^n-a1)/(q-1)*[a1q^(n+2)-a1]/(q-1)=a1²/(q-1)²*(q^n-1)(q²*q^n-1)
S²n+1=[(a1q*q^n-a1)/(q-1)]²=a1²/(q-1)²*(q*q^n-1)²
(q^n-1)(q²*q^n-1)-(q*q^n-1)²=[q^(2n+2)-q^n-q^(n+2)+1]-[q^(2n+2)-2q^(n+1)+1]
=q^(2n+2)-q^n-q^(n+2)+1-q^(2n+2)+2q^(n+1)-1
=q^n*(2q-1-q²)
=-q^n*(q-1)²<0
∴(q^n-1)(q²*q^n-1)<(q*q^n-1)²
∴Sn*Sn+2<S²n+1
∵0<a<1
∴loga Sn*Sn+2>loga S²n+1
logaSn+logSn+2>2logaSn+1
∴1/2(logaSn+logaSn+2)>logaSn+2
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