√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2

题目

√[1+1/(1^2)+1/(2^2)] = 1+1/1-1/2=1+1/2
√[1+1/(2^2)+1/(3^2)] = 1+1/2-1/3=1+1/6
√[1+1/(3^2)+1/(4^2)] = 1+1/3-1/4=1+1/12
根据上述等式揭示的规律,写出用字母n (n为大于1)的自然数表示这一规律的等式,并说明等式成立的理由

答案

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))