1.设等差数列an，bn的前n项和分别为Sn，Tn若对任意自然数n都有Sn/Tn=2n-3/4n-3

1.设等差数列an，bn的前n项和分别为Sn，Tn若对任意自然数n都有Sn/Tn=2n-3/4n-3

a9/(b5+b7)+a3/(b8+b4)

=a9/2b6+a3/2b6

=a9+a3/2b6

=2a6/2b6

=(a1+a11)/(b1+b11)

=11(a1+a11)/11(b1+b11)

=[11(a1+a11)/2]/[11(b1+b11)/2]

=S11/T11

=19/41

解：

设数列{an}公差为d，数列{bn}公差为d'。

Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)d'/2]

=[dn+(2a1-d)]/[d'n+(2b1-d)]=(2n-3)/(4n-3)

令d=2t，则2a1-d=-3t d'=4t 2b1-d=-2t

解得a1=-t/2 d=2t b1=t d'=4t

a9/(b5+b7)+a3/(b8+b4)

=a9/(2b6)+a3/(2b6)

=(a3+a9)/(2b6)

=(2a6)/(2b6)

=a6/b6

=(a1+5d)/(b1+5d')

=(-t/2 +10t)/(t+20t)

=95/42

a9/(b5+b7)+a3/(b8+b4)

=a9/2b6+a3/2b6

=a9+a3/2b6

=2a6/2b6

=(a1+a11)/(b1+b11)

=11(a1+a11)/11(b1+b11)

=[11(a1+a11)/2]/[11(b1+b11)/2]

=S11/T11

=19/41

b5+b7=b8+b4=2b6，T11=11b6，a9/(b5+b7)+a3/(b8+b4)=11*（a9+a3）/(2T11)

又a9+a3=2a6,即S11=11a6，所以原式=S11/T11=19/41