已知等差数列前n项和分别是SnTn且Sn/Tn=2n/3n+1则a2+a4+a6/b2+b4+b6=?

Question

已知等差数列前n项和分别是SnTn 且Sn/Tn=2n/3n+1则 a2+a4+a6/b2+b4+b6=?

Answer 1

S1/T1=a1/b1=2/(3+1)=1/2
b1=2a1
S2/T2=(2a1+d1)/(2b1+d2)=4/7
8b1+4d2=14a1+7d1
b1=2a1代入,解得a1=(7d1-4d2)/2
S3/T3=(3a1+3d1)/(3b1+3d2)=(a1+d1)/(b1+d2)=6/10=3/5
3b1+3d2=5a1+5d1
b1=2a1代入,解得a1=5d1-3d2
(7d1-4d2)/2=5d1-3d2
3d1=2d2
d1/d2=2/3 a1=5d1-(9/2)d1=d1/2 b1=d1
(a2+a4+a6)/(b2+b4+b6)
=(3a1+9d1)/(3b1+9d2)
=[(3/2)d1+9d1]/[3d1+9(3/2)d1]
=21/15
=7/5