已知等比数列{an}共有n+1项,其首项a1=1,末项a(n+1)=2002,公比q＞0(1)记Tn=a1a2a3...an,求Tn(2)记bn=t^log(2002)Tn(t＞0且t≠1),求数列{bn}的前n项之和Sn(3)在(2)的条件下,试比较Sn/bn与S(n+1)/b(n+1)的大小

Question

已知等比数列{an}共有n+1项,其首项a1=1,末项a(n+1)=2002,公比q＞0 (1) 记Tn=a1a2a3...an,求Tn
(2) 记bn=t^log(2002)Tn(t＞0且t≠1),求数列{bn}的前n项之和Sn
(3) 在(2)的条件下,试比较Sn/bn与S(n+1)/b(n+1)的大小

Answer 1

a1=1,a2=q ,an=q^(n-1) a2*an=a(n+1)=2002
Tn=a1a2a3...an=2002^[(n-1)/2]
bn=t^log(2002)Tn=t*(n-1)/2=n*(t/2)-t/2 ,为首项0,公差为t/2的等差数列
Sn=t*(n-1)*n/4
Sn/bn=[t*(n-1)*n/4]/[t*(n-1)/2]=n/2
S(n+1)/b(n+1)=(n+1)/2
Sn/bn＜S(n+1)/b(n+1)