规律：√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))
证明：假设√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))成立
则：（1+1/(n+1)-1/(n+2))^2=1+2/(n+1)-2/(n+2)+1/((n+1)^2)+1/((n+2)^2)-2/((n+1)(n+2))
=1+1/((n+1)^2)+1/((n+1)^2)
即：1+1/((n+1)^2)+1/((n+1)^2)=（1+1/(n+1)-1/(n+2))^2
两边各开根号：得：√1+1/((n+1)^2)+1/((n+1)^2)=1+1/(n+1)-1/(n+2)
又n (n为大于1)的自然数,n=1时等式成立
综上所述：√[1+1/(n^2)+1/((n+1)^2)] = 1+1/n-1/(n+1)=1+1/(n*(n+1))