Question

等差数列｛an｝，{bn}的前n项和为Sn，Tn。若对任意n∈N*，都有Sn/Tn=4n+3/2n

Answer 1

an/bn=S(2n-1)/T(2n-1)
=[4(2n-1)+3]/[2(2n-1)-3]
=(8n-1)/(4n-5)
=(8n-10+9)/(4n-5)
=2
+9/(4n-5)
要an/bn为正整数，9/(4n-5)为正整数
n只能是2
满足题意的n只有一个，n的值为2

Answer 2

解：
设等差数列{an}的公差为a
等差数列{bn}的公差为b
s(2n-1)=［a1+a(2n-1)］*(2n-1)/2
=2an*(2n-1)/2
=(2n-1)an
t(2n-1)=［b1+b(2n-1］*(2n-1)/2
=2bn*(2n-1)/2
缉础光飞叱读癸嫂含讥
=(2n-1)bn
s(2n-1)/t(2n-1)=［(2n-1)an］/［(2n-1)bn］
=an/bn
∴an/bn=[4(2n-1)+27]/{7(2n-1)+1]=(8n+23)/(14n-6)