根号下（1+1平方分之一+2平方分之一)等于?有何规律?

√(1+1/1²+1/2²)=√(9/4)=3/2=1+1/2
你就写一个式子,那有什么规律呀
再写几个如：
√(1+1/2²+1/3²)=√(2²·3²+2²+3²)/(2²3²)=7/(2×3)=1+1/6
√[1+1/n²+1/(n+1)²]
=1/[n(n+1)]* √[n²(n+1)²+(n+1)²+n²]
=1/[n(n+1)]* √[n²(n+1)²+2n(n+1)+1]
=[n(n+1)+1]/[n(n+1)]
=1+1/[n(n+1)]
=1+1/n-1/(n+1)
∴√(1+1/1²+1/2²)+√(1+1/2²+1/3²)+.+√[1+1/n²+1/(n+1)²]
=(1+1-1/2)+(1+1/2-1/3)+.+[1+1/n-1/(n+1)]
=n+[1-1/2+1/2-1/3+.+1/n-1/(n+1)]
=n+1-1/(n+1)
=n+n/(n+1)