an+an-1=(n/(an-an-1))+2 去分母整理得 （an-1）^2=(a(n-1)-1)^2+n
=>
（an-1）^2=(a(n-1)-1)^2+n
=(a(n-2)-1)^2+n-1 +n
...
= 1+2+ ...+n = n(n+1)/2
1/（an-1）^2 = 1/(n(n+1)/2) = 2/n - 2/(n+1)
==>
1/(a1-1)^2+1/(a2-1)^2+...+1/(an-1)^2 = 2 - 2/(n+1) --> 2