Question

已知数列{an}和{bn}都是等差数列,他们的前n项和分别记为Sn和Tn,且Sn/Tn=6n/(2n＋3）

求a5／b5

Answer 1

a5/b5
=2a5/2b5
=(a1+a9)/(b1+b9)
=9(a1+a9)/9(b1+b9)
=[9(a1+a9)/2]/[9(b1+b9)/2]
=S9/T9
=54/21
=18/7

追问

2a5/2b5
=(a1+a9)/(b1+b9)这是为什么啊

追答

等差数列性质
m+n=p+q
am+an=ap+aq

Answer 2

Sn/Tn=6n/(2n＋3）
S(2n-1)/T(2n-1)=6(2n-1)/[2*(2n-1)+3]
{[a1+a(2n-1)]*(2n-1)/2}/{[b1+b(2n-1)]*(2n-1)/2}=(12n-6)/(4n+1)
[a1+a(2n-1)]/[b1+b(2n-1)]=(12n-6)/(4n+1)
an/bn=(12n-6)/(4n+1)

a5/b5
=(12*5-6)/(4*12+1)
=54/49