Question

高中数学 数列题帮我解答一下，谢谢，题目如下

Answer 1

an= 2-1/a(n-1)
an-1 = 1- 1/a(n-1)
= [a(n-1) -1]/ a(n-1)
1/(an-1) = a(n-1) /[a(n-1) -1]
= 1+ 1/[a(n-1) -1]
1/(an-1) - 1/[a(n-1) -1] =1
1/(an-1) -1/(a1-1) = n-1
1/(an-1) =(2n- 7)/2
an-1 = 2/(2n-7)
an = (2n-5)/(2n-7)

Answer 2

证明：令bn=1/(an-1)(n∈N*)
an*a(n-1)+1=2a(n-1)
an=[2a(n-1)-1]/a(n-1)

an -1= [2a(n-1)-1]/a(n-1) -1= [a(n-1)-1]/a(n-1)
1/(an -1) = a(n-1)/[a(n-1)-1] = {[a(n-1)-1]+1} /[a(n-1)-1]

=1+ 1/[a(n-1) -1]
1/(an -1) -1/[a(n-1) -1] =1

即 bn － b(n-1)= 1

所以bn 是等差数列。
bn =b1 +1*(n-1)= 1/(a1 -1) +n-1=1/(-2/5) +n-1= n-7/2, n>=2

an =1/bn +1
=1/(n-7/2) +1
= 1+ 2/(2n-7)

Answer 3

an-1=1-1/(a(n-1)=(a(n-1)-1)/(a(n-1) )两边倒数得到
1/（an-1）=(a(n-1))/（a(n-1)-1)=1+1/（a(n-1)-1)
把1/（an-1）当做一整体1/（an-1）是个等差数列又1/（a1-1）=-5/2
所以1/（an-1）=n-7/2
an=1+2/(2n-7)=(2n-5)/(2n-7)