设等比数列{q^n-1}(|q|>1)的前n项和为Sn,则Sn+2/Sn的极限是

来源：学生作业帮编辑：搜狗做题网作业帮分类：数学作业时间：2024/04/29 19:07:33

设等比数列{q^n-1}(|q|>1)的前n项和为Sn,则Sn+2/Sn的极限是
A.1/q² B.1/q四次方 C.q² D.q四次方

$设等比数列{q^n-1}(|q|>1)的前n项和为Sn,则Sn+2/Sn的极限是$

Sn=(1-q^n)/(1-q)
所以sn+2=(1-q^(n+2))/(1-q)
所以sn+2/sn=(1-q^(n+2))/(1-q^n)=q^2
即c

设等比数列{an}的公比为q，前n项和为Sn，若Sn+1，Sn，Sn+2成等差数列，则公比q为（　　）设等比数列｛an｝的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q=? 设等比数列｛an｝的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q等于多少?若an=1,求sn前n 设等比数列｛an｝的公比为q,前n项和为Sn,若S(n+1),Sn,S(n+2)成等差数列,则q= 设等比数列{An}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则q的值为（）我知道答案是-2,可设等比数列an的公比为q,前n项和为sn,若s(n+1),sn,s(n+2)成等差数列,求q的值已知数列{an}是首项为a1,公比为q(q>0)的等比数列,前n项和为sn,求(sn/(sn+1))的极限我就想问一设等比数列{ an}的公比为q,q>0且q≠1,Sn为{an}的前n项和,记Tn=an/Sn,则设Sn为等比数列{an}的前n项和,已知Sn=3an+1+m,Sn-1=3an+m,则公比q= 设等比数列an的公比为q=1/2,前n项和为Sn,则S4/a4=? 已知等比数列{an}的公比为q,前n项和为Sn,求[Sn*Sn+2-（Sn+1）^2]/[an*an+2] 设{an}是等比数列,公比q=根号2,Sn为{an}的前n项和,记Tn=（17Sn-S2n）/an+1