(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,...
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,...
An=A1+(n-1)d 等差数列的前n项和: Sn=[n(A1+An)]/2; Sn=nA1+[n(n-1)d]/2 等差数列求和公式:等差数列的和=(首数+尾数)*项数/2;等比数列 : 通项公式: an=a1×q^(n-1); 等比数列的前n项和: Sn=n×a1 (q=1) Sn=a1(1-q^n)/(1-q) =(a1-an×q)/(1-q) (q≠1) 解析看...
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: ...
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...
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 ...
19.等差数列{an}的前n项和为Sn,数列{bn}是等比数列,满足a1=3,b1=1,b2+S2=10,a5﹣2b2=a3.(1)求数列{an}和{bn}的通项公式;(
(1)∵Sn=2n2-3n,∴当n=1时,a1=S1=2-3=-1,当n≥2时,an=Sn-Sn-1=(2n2-3n)-[2(n-1)2-3(n-1)]=4n-5,由于a1也适合此等式,∴an=4n-5.(2)∵Sn=3n+b,∴当n=1时,a1=S1=3+b,当n≥2时,an=Sn-Sn-1=(3n+b)... (1)由Sn=2n2-3n,利用公式 an= S 1,n=1 Sn−Sn−1,...
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，首项a 1 = 3，公差d = 2，此时等差数列{a n }前n项的和S n = n*3 + n(n –1)d/2 = 3n + n 2 –n = n 2 + 2n，b = ...