Question

请教数学题！

1.设数列{an}和数列{bn}的前n项和为Sn,Tn，且Sn/Tn=7n+2/n+3 ,则a8/b8=？
2.在边长为R正六边形内，依次连接各边中点得一个正六边形，又在所得正六边形内再依次连接中点得正六边形……这样无限继续下去。设前N个正六边形的边长之和为Sn，所有这些正六边形的边长之和为S,则S=？

Answer 1

(1)
Sn/Tn=(7n+2)/(n+3)
Sn/(7n+2)=Tn/(n+3) =k
Sn =(7n+2)k , Tn= (n+3)k
a8/b8
=(S8-S7)/(T8-T7)
=(58k-51k)/(11k-10k)
=7
(2)
let an =第N个正六边形的边长之和
a1=6R
(a2/6)^2 = (R/6)^2 + (R/6)^2 - 2(R/6)(R/6)cos(2π/3)
= 3R^2/36
a2 = √3R
...
...
(an/6)^2 = (a(n-1)/6)^2 + (a(n-1)/6)^2 - 2(a(n-1)/6)(a(n-1)/6)cos(2π/3)
= 3a(n-1)^2/36
an= √3a(n-1)
an/a(n-1) = √3
an/a1= (√3)^(n-1)
an= 6R (√3)^(n-1)

Sn = a1+a2+...+an
= 6R[(√3)^n -1] /(√3-1 )

Answer 2

一定存在m,使Sn=m（7n+2）,Tn=m(n+3)
a8=S8-S7=m(58-51)=7m
b8=T8-T7=m(11-10)=m
a8/b8=7

边长和后一个是前一个的√3/2
an=6R*（√3/2）^n
Sn=a1(1-q^n)/(1-q)
=3√3R(1-（√3/2）^n)/(1-√3/2)
当n趋向于无穷大时，（√3/2）^n=0
所有这些正六边形的边长之和
S=3√3R/(1-√3/2)
=（12√3+18)R