(2)由a2=4,S4=20,可得3k+b=4,16k+4b=20,解得k=b=1,即an=2n,由b1=6,且(b_(n+1))/(b_n)=(a_n)/(a_(n+2))=(2n)/(2(n+2))=n/(n+2),则bn=b1•(b_2)/(b_1)•(b_3)/(b_2)•...•(b_(n-1))/(b_(n-2))•(b_n)/(b_(n-1))=6×1/3×2...
AW01,TS01S,TS02N,TS04N,TS04P,TS06,TS08N,TS08P,TS12,TS20,TSM12S,TSM12MC,TSM16,ANSG08SH,ANMG08PL,ANMG04SG,ANSG08SL,ALRS08FB; ★JOULWATT(杰华特)热销产品: JW5250A,JW5250S,JW5211,JW5033S,JW5018,JW5018B,JW5026,JW5255A,JW5255,JW3651,JW3655E,JW5361S,JW5361M,JW5116F,JW5062,...
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: ...
1若两个等差数列的前n项和分别为An、Bn,且满足,则的值为( ) 2(5分)若两个等差数列{an},{bn}的前n项和分别为An、Bn,且满足n2n-1B3n+1n,则a3+a7+a11b5+bg的值为( ) A. 3944 B. 58 C. 1516 D. 1322 3(5分)若两个等差数列{an}、{bn}的前n项和分别为An、Bn,且满足n4n+2B5n-5n,...
数列{an}的前n项和为Sn,an是Sn和1的等差中项,等差数列{bn}满足b1+S4=0,b9=a1.(1)求数列{an}和{bn}的通项公式;(2)若cn=1(bn+16)(bn+18),Wn是数列{cn}的前n项和,求Wn及取值范围。 答案 (1)由于an是Sn和1的等差中项,则2an=1+Sn,①当n=1时,2a1=1+S1=1+a1,解得,a1=1,又n...
所以,an=1+(n-1)×1=n,(S_n)=((n(n+1)))/2;(Ⅱ)设bn=b1+(n-1)d,b2=12,b5=30,即b1+d=12,b1+4d=30,解得b1=6,d=6,所以,bn=6n,由kSn≥bn得:k≥((12n))/((n(n+1)))=((12))/((n+1)),设(c_n)=((12))/((n+1)),则{cn}是递减数列,...
解答一 举报 sn/tn=(7n+2)/(n+3) Sn/Tn=n(7n+2)/n(n+3)Sn/Tn=[n(7n+2)/2]/[n(n+3)]/2Sn=n(a1+an)/2Tn=n(b1+bn)/2a1+an=7n+2b1+bn=n+3a7/b7=2a7/2b7=(a1+a13)/(b1+b13)=(7*13+2)/(13+3)=93/16 解析看不懂?免费查看同类题视频解析查看解答 ...
S11<S10 答案 D【分析】由题意可得数列的前8项为负数,第9项为0,从第10项开始为正数,各个选项验证可得答案.相关推荐 1设数列{an}是等差数列,Sn是{an}的前n项和,且S7>S8,S8=S9<S10,则下列结论错误的是( ) A. d>0 B. a9=0 C. S8,S9均为Sn的最小值 D. S11<S10 反馈 收藏 ...
解析 解:由an=2n-1,得Sn=a1+a2+…+an=(21-1)+(22-1)+(23-1)+…+(2n-1)=(21+22+23+…+2n)-n=2(1-2”) -n=2n+1-2-n 1-2.故选:A. 结果一 题目 若数列{an}的通项公式为an=2n-1,则数列{an}的前n项和Sn等于( ) A. 2n+1-n-2 B. 2n+1-n C. 2n+1-n+2 D. 2n+1...
【解答】(1)证明:∵Sn为数列{an}的前n项和,Sn=nan-3n(n-1)(n∈N*),且a2=11.∴S2=a1+a2=2a2-3×2(2-1),解得a1=5,当n≥2时,Sn-1=(n-1)an-1-3(n-1)(n-2),由an=Sn-Sn-1,得an=nan-3n(n-1)-(n-1)an-1-3(n-1)(n-2),∴(n-1)an-(n-1)an-1=6(n-1),∴...