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又因为 Sn+m−1=a1+an+m−12(n+m−1) ,所以由 an+am>0 能得到 Sn+m−1>0 题1: {an} 为等差数列, a2003>0,a2004<0,a2003+a2004>0 ,求使得 Sn>0 成立的最大 n 值为( ) A.4005 B.4006 C.4007 D.4008 极简分析:利用结论: ...
等差数列{An},{Bn}的前n项和为Sn与Tn,若Sn/Tn=2n/3n+1,则An/Bn的值是? {an},{bn}均为等差数列,前n项和分别为Sn,Tn且anbn=3n+74n+1,则S11T11=( ) A.2221 B.1 C.89 D.1417 已知等差数列{an},{bn}满足an/bn=2n/(3n+5),它们的前几项之和分别记为Sn和Tn,求S11/T11的值(要过程) ...
解：设等差数列{a n }的公差为d，前n项的和为S n ，所以S n = na 1 + n(n –1)d/2 = (d/2)n 2 + (a 1 –d/2)n = an 2 + bn，所以 b = a 1 –d/2 （b是等差数列{a n }的首项a 1 与公差的一半 之差）；举例：对于等差数列{a n }：a n = 2n + 1，首...
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当n≥2时an=Sn-S(n-1)=[An^2+Bn]-[A(n-1)^2+B(n-1)]=2An+(B-A)当a1=s1=A+B,满足an=2An+(B-A)所以an=2An+(B-A)是数列{an}的通项公式,首项是a1=A+B又公差d=a(n+1)-an=2A(n+1)+(B-A)-2An-(B-A)=2A,所以{an}是等差数列,... 解析看不懂?免费查看同类题视频解析...
已知两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且=,则为( ) A. \u2220A B. C. D. 相关知识点: 试题来源: 解析 解:由题意和等差数列的求和公式和等差数列的性质可得:===2 9(b1+b9) 2===故选:C 结果一 题目 已知两个等差数列{an}和{bn}的前n项和分别为Sn和Tn,且S.n T,=2n+5n...
SN74LS47N 原装正品 SN74LS系列逻辑IC BCD7段解码器驱动芯片 安装类型 SN74LS123NG4、SN74LS540NSR、SN74LS641-1DW、SN74LS74ADG4、SN74LS125AD、SN74LS123DBR、SN74LS541DW、SN74LS244NS、SN74LS221NE4、SN74LS367ADR、SN74LS240NS、SN74LS367AN、SN74LS540DWR、SN74LS123DR、SN74LS221NS、SN74...
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indicating the formation of an adduct between the carbonyl oxygen and the Sn incorporated in the framework of the zeolite (Figure 1a). With Si-beta or Si-beta impregnated with Sn no displacement of the carbonyl signal was obtained (Figure 1b). This clearly demonstrated that the Sn incorporated...