已知{an}和{bn}均为等差数列,Sn和Tn分别表示它们的前n项和,若 Sn/Tn =(4n-2)/2n+1),则a6/b6的值为: A 22/13 B 10/7 C 46/25 D 42/23
热心网友
以上的答案都不好，我来做一个精彩的答案∵a6/b6=(2a6)/(2b6)=(a1+a11)/(b1+b11)=［11（a1+a11）/2］/［11（b1+b11）/2］=S11/T11=(4*11-2)/(2*11+1)=42/23∴选D。
热心网友
因为 Sn/Tn=(4n-2)/(2n+1)所以 Sn=kn(4n-2); Tn=kn(2n+1)因此 a6/b6=(S6-S5)/(T6-T5)=(6*22k-5*18k)/(6*13k-5*11k)=(132-90)/(78-55)=42/23.故选 D.
热心网友
a1/b1=Sn/Tn=2/3 a1=(2/3)b1 (1) 把n=1代入S2/T2=(a1+a2)/(b1+b2)=(2a1+d1)/(2b1+d2)=6/5(2)把n=2代入S3/T3=(3a1+2d1)/(3b1+2d2)=10/7 (3) 把n=3代入联立(1)(2)(3)解得d1=(2b1)/3 d2=-b1/3a6/b6=(6a1+5d1)/(6b1+5d2)=22/13故选A
热心网友
c
热心网友
已知{an}和{bn}均为等差数列,Sn和Tn分别表示它们的前n项和,若 Sn/Tn =(4n-2)/(2n+1),则a6/b6的值为: A 22/13 B 10/7 C 46/25 D 42/23 a1/b1 = S1/T1 = 2/3……………………………………………………………①(a1 + a2)/(b1 + b2) = (2a1 + d1)/(2b1 + d2) = S2/T2 = 6/5…………②(a1 + a2 + a3)/(b1 + b2 + b3) = (3a1 + 3d1)/(3b1 + 3d2) = S3/T3 = 10/7 …………………………………………………………………③(a1 + a2 + a3 + a4)/(b1 + b2 + b3 + b4) = (4a1 + 6d1)/(4b1 + 6d2) = S4/T4 = 14/9 …………………………………………………………………④联立①、②、③、④解得b1 = 3a1/2，d1 = 4a1 d2 = 2a1a6/b6 = (6a1 + 15d1)/(6b1 + 15d2) = (6a1 + 60a1)/(9a1 + 30a1) = 66/39 = 22/13选A。。