设等差数列{an}{bn}的前几项的和分别为Sn,Tn,若Sn/Tn=n/2n+1,则a5/b5=?,a5/b3=?

Question

设等差数列{an}{bn}的前几项的和分别为Sn,Tn,若Sn/Tn=n/2n+1,则a5/b5=?,a5/b3=?

Answer 1

已知等差数列{an}{bn}的前几项的和分别为Sn,Tn
设公差分别为d和d'
Sn=[2a1+(n-1)d]*n/2 Tn=(2b1+(n-1)d']*n/2
已知Sn/Tn=n/(2n+1)
当n=1时 a1/b1=S1/T1=1/3 b1=3a1 (1)
当n=21时 S2/T2=(2a1+d)/(2b1+d')=(2a1+d)/(6a1+d')=2/5
2(6a1+d')=5(2a1+d)
12a1+2d'=10a1+5d
解得d'=5d/2-a1 (2)
当n=3时 Sn/T3=(2a1+2d)/(2b1+2d')=(a1+d)/(b1+d)=3/7
(1)(2)代入 (a1+d)/(2a1+5d/2)=3/7
6a1+15d/2=7a1+7d
解得d=2a1 (3)
于是d'=5d/2-a1=4a1 (4)
所以a5/b5=(a1+4d)/(b1+4d')=(a1+8a1)/[3a1+16a1)] [(1)(3)(4)代入]
=(9a1)/(19a1)
=9/19
a3/b3=(a1+2d)/(b1+2d')=(a1+4a1)/(3a1+8a1) [(1)(3)(4)代入]
=(5a1)/(11a1)
=5/11