已知数列{an}满足a1=3,an=2-（1/an-1）.求证：数列{1/an-1}是等差数列,并写出{an}的一个通项

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已知数列{an}满足a1=3,an=2-（1/an-1）.求证：数列{1/an-1}是等差数列,并写出{an}的一个通项公式.
注明：n、n-1都是a的下标.
没有看明白过程.请不要粘贴.

an=2- 1/a(n-1）
= [2a(n-1) -1]/a(n-1)
an -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) }是等差数列, d=1
1/( an -1) -1/( a1 -1) = n-1
1/( an -1) = (2n+1)/2
an = 2/(2n+1) +1
= (2n+3)/(2n+1)

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