已知{An}是首项为a1,公比为q,(q不等于1,大于0)的等比数列,前n项和为Sn,5*S2=4*S4,设Bn=q+Sn (1)
已知{An}是首项为a1,公比为q,(q不等于1,大于0)的等比数列,前n项和为Sn,5*S2=4*S4,设Bn=q+Sn (1)求q (2)数列Bn能否为等比数列.若能求q,不能说出理由.麻烦写出过程,
人气:161 ℃ 时间:2019-10-24 01:39:26
解答
5*(1-q^2)/(1-q)=4*(1-q^4)/(1-q)去掉分母,解关于q^2的一元二次方程5-5*(q^2)=4-4*(q^2)^24*(q^2)^2-5*(q^2)+1=0(q^2)=1或1/4,因为q不等于1,大于0,所以把1舍去,q^2=1/4,q=1/2.B1=1.5B2=2.0B3=2.25B4=2.375...Bn不是...
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