数列问题8

一。根据
Bn=An-1/2 得到
B[n+1]=A[n+1]-1/2
把An替换成Bn (1-An)*A[n+1]=1/4
得到
（1/2-Bn)*(B[n+1]+1/2)=1/4
1/4+1/2(B[n+1]-B[n])-B[n]*B[n+1]=1/4
B[n+1]-B[n]=2*B[n]*B[n+1]
等是两边同时除以B[n]*B[n+1]
得到
1/B[n]-1/B[n+1]=2
1/B[n+1]=1/B[n]-2
所以数列（1/B[n]）为以-2为公差的等差数列
因为B[1]=A[1]-1/2=-1/4
1/B[1]=-4
所以(1/B[n])=-4-2(n-1)=-2n-2=-2(n+1)
所以B[n]=-1/[2(n+1)]

二、
A[n]=B[n]+1/2=-1/[2(n+1)]+1/2=n/[2(n+1)]
A[n+1]/A[n]=[(n+1)*(n+1)]/[(n+2)*n]=
A2/A1+A3/A2+……+A(n+1)/An
=∑{1+1/[(n+2)*n]}
=n+∑1/[(n+2)*n]
=n+1/2*∑[1/n-1/(n+2)]
=n+1/2*[1+1/2-1/(n+1)-1/(n+2)]
=n+3/4-1/2*[1/(n+1)+1/(n+2)]<n+3/4 (n为正整数）
证毕

(1-An)*A(n+1)=1/4
即
（1/2+1/2-An)[1/2-(1/2-A(n+1)]=1/4
又因为Bn=An-1/2 ，则
(1/2-Bn)[1/2+B(n+1)]=1/4
化简：
1/2*B(n+1)-1/2*Bn-B(n+1)*Bn=0
同时除以1/2*B(n+1)*Bn
1/Bn-1/B(n+1)=2
1/Bn=1/B1-2*(n-1)Bn=1/[1/B1-2*(n-1)]

B1=-1/4
Bn=1/(-4-2n+2)=