1-2a(n-1)+a(n-1).an=0
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)
= 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) =n
an = 1 + 1/n
=(n+1)/n
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