怎样简便计算?（√1+1/1²+1/2²）+（√1+1/2²+1/3²）+（√1+1/3²+1/4²）+...+（√1+1/99²+1/100²）

怎样简便计算?
（√1+1/1²+1/2²）+（√1+1/2²+1/3²）+（√1+1/3²+1/4²）+...+（√1+1/99²+1/100²）

S=√(1+1/1²+1/2²)+√(1+1/2²+1/3²)+√(1+1/3²+1/4²)+...+√(1+1/99²+1/100²)
通项an=√[1+1/n²+1/(n+1)²]
=√{[(n+1)²n²+(n+1)²+n²]/[n²(n+1)²]}
=1/[n(n+1)]* √[n²(n+1)²+2n²+2n+1]
=1/[n(n+1)]*√[n²(n+1)²+2n(n+1)+1]
=1/[n(n+1)]*√[n(n+1)+1]²
=[n(n+1)+1]/[n(n+1)]
=1+1/[n(n+1)]
=1+1/n-1/(n+1)
∴S=a1+a2+a3+.+a99
=(1+1-1/2)+(1+1/2-1/3)+(1+1/3-1/4)+.+(1+1/99-1/100)
=99+(1-1/2+1/2-1/3+1/4-1/4+.+1/99-1/100)
=99+1-1/100
=99又99/100