各项都是正数的等比数列的公比不等于1， a3,a5,a6成等差数列，则a3+a5与a4+a6的比是多少 解:设各项都是正数的等比数列的首项为a1,公比q≠1, →a3=a1q^2,a5=a1q^4,a6=a1q^5(^2表平方,^4表4方,^5表5次方) a3,a5,a6成等差数列，→a3+a6=2a5,→ a1q^2+a1q^5=2a1q^4,→a1q^2(1+q^3)=2a1q^4,→ (1+q^3)=2q^2,→1-2q^2+q^3,→1-q^2-q^2+q^3=0,→ (1-q^2)-(q^2-q^3)=0,→(1-q)(1+q)-q^2(1-q)=0,→ (1-q)[(1+q)-q^2]=0,(1-q)≠0→ (1+q)-q^2=0,→q^2-q-1=0,求根公式: q=(1±√5)/2,∵各项都是正数,q>0 ∴q=(1+√5)/2 ∴(a3+a5)/(a4+a6)=a1q^2(1+q^2)/a1q^3(1+q^2) =1/q=2/(1+√5)=2(√5-1)/(√5+1)(√5-1))= 2(√5-1)/4=(√5-1)/2。
设等比数列的公比是q an = a1*q^(n-1) a3 = a1*q*2; a4 = a1*q^3; a5 = a1*q^4; a6 = a1*q^5; a3, a5, a6成等差数列； a3 + a6 = 2a5; a1*q^2 + a1*q^5 = 2a1*q^4; a1>0, q>0; 1+q^3 = 2q^2; q^3 - 2q^2 + 1 = 0; (q-1)(q^2-q-1) = 0; q = 1, q = (1+sqrt(5))/2, q = (1-sqrt(5))/2; 由于各项都是正数的等比数列的公比不等于1， 故 q = (1+sqrt(5))/2; a3+ a5 = a1*q^2+a1*q^4 = a1*q^2(1+q^2); a4+a6 = a1*q^3 +a1*q^5 = a1*q^3(1+q^2); (a3+a5)/(a4+a6) = 1/q = 2/(1+sqrt(5)) =2*(sqrt(5)-1)/(5-1) = (sqrt(5)-1)/2。
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答：等差： Sn=（a1+an）×n/2=a1×n+n(n-1)d/2=An＾+Bn 其中a1是数列首项,d是公差. A=d/2 B=a1-(d/2) 等...详情>>
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