設(shè)公比為q
a1+1,2s2,3s3 成等差數(shù)列
則a1+1+3S3=2*(2S2)=4S2
即a1+1+3a1(1-q^3)/(1-q)=4a1(1-q^2)/(1-q)
a1+1+3a1(1+q+q^2)=4a1(1+q)
3a1*q^2-a1*q+1=0
若存在唯一實(shí)數(shù)q,即只有一個(gè)根
則(-a1)^2-4*3a1*1=0
a1=12
代入方程36q^2-12q+1=0 q=1/6
an=a1*q^(n-1)=12/6^(n-1)=2/6^(n-2)
等比數(shù)列{an}首項(xiàng)為a1前n項(xiàng)和為Sn,若存在唯一實(shí)數(shù)q使 a1+1,2s2,3s3 成等差數(shù)列,則｛an｝的通項(xiàng)=
等比數(shù)列{an}首項(xiàng)為a1前n項(xiàng)和為Sn,若存在唯一實(shí)數(shù)q使 a1+1,2s2,3s3 成等差數(shù)列,則｛an｝的通項(xiàng)=
數(shù)學(xué)人氣：490 ℃時(shí)間：2020-09-06 09:18:32
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