高中数列加试题

求所有正整数A1,A2...An,使得(99/100)=(A0/A1)+(A1/A2)+...+(An-1/An),其中A0=1,[A(k+1)-1]*A(k-1)>=A(k)*A(k)*[A(k)-1],k=1,2,...n
请写出具体过程，不胜感激

[A(k+1)-1]*A(k-1)>=A(k)*A(k)*[A(k)-1]
(a2-1)*a0=a1^2(a1-1)
a0/a1=a1(a1-1)/(a2-1)=a1a2(a1-1)/a2(a2-1)=(a1/a2)*[a2(a1-1)/(a2-1)]

a1/a2=(a0/a1)*(a2-1)/a2(a1-1)

a2/a3=(a1/a2)*(a3-1)/a3(a2-1)=(a0/a1)*(a3-1)/a2a3(a1-1)

a3/a4=(a2/a3)*(a4-1)/a4(a3-1)=(a0/a1)*(a4-1)/a2a3a4(a1-1)

...

a(n-1)/an=[a(n-2)/a(n-1)]*(an-1)/an[a(n-1)-1]
=(a0/a1)*(an-1)/a2a3...an(a1-1)

1/a2a3...an=[a(n-1)/an]*(a1/a0)*(a1-1)/(an-1)

a0/a1+a1/a2+...+a(n-1)/an
=(a0/a1)/(a1-1)[a1-1+(a2-1)/a2+(a3-1)/a2a3+...+(an-1)/a2a3...an]
=a0/[a1(a1-1)]*[a1-1+1-1/a2+1/a2-1/a2a3+1/a2a3-1/a2a3a4+...+1/a1a2...a(n-1)-1/a2a3...an]
=a0/[a1(a1-1)]*[a1-1/a2a3...an]
=[1/(a1-1)][(a1a2...an-1)/(a1a2...an)]
=99/100

令a1a2...an=A

(a1-1)*[A/(A-1)]=100/99

a1=100(A-1)/99A+1=(199A-100)/99A,为正整数，设为N
(199A-100)/99A=N
A=100/(199-99N) N只能为1,2,

N=1时，A=1 又a0=1,因此a1=a2=...an=1,代入，不符合题意。