高中数学数列题目，给出答案和详细过程，谢谢了。

Question

如题所述

Answer 1

a1=1/4
a2=1/3=2/6
a3=3/8
a4=2/5=4/10
……
an=n/[2*(n+1)]
a(n+1)=1/[4-4*an]=(n+1)/[2(n+1+1)]成立
a(n+1)/an=(n+1)^2/[n*(n+2)]
=1+1/[n*(n+2)]
=1+1/2*[1/n-1/(n+2)]
所以有：a2/a1+a3/a2+a4/a3+……++a(n+1)/an+a(n+2)/a(n+1)
=1+1/2[1/1-1/3]+1+1/2[1/2-1/4]+1+1/2[1/3-1/5]+……+1+1/2[1/n-1/(n+2）+1+1/2[1/(n+1)-1/(n+3)]
=n+1+1/2[1+1/2+……+1/(n+1)-[1/3+1/4+1/5+……+1/(n+3)]]
=n+1+1/2*[1+1/2-1/(n+2)-1/(n+3)]
=n+1+1/2*[3/2-(2n+5)/[(n+2)(n+3)]
=n+7/4-(2n+5)/[2*(n+2)*(n+3)]<n+7/4
选D追问

n+1+1/2[1+1/2+……+1/(n+1)-[1/3+1/4+1/5+……+1/(n+3)]],n是怎么得来的。

Answer 2

a2/a1>1,a3/a2>1
a2/a1+a3/a2>2<1+A
A>1
选D，7/4