就用这个方法--
化简√﹙1+1/1²+1/2²﹚+√﹙1+1/2²+1/3²﹚+√﹙1+1/3²+1/4²﹚+……+√﹙1+1/2012²+1/2013²﹚
∵√﹙1+1/1²+1/2²﹚=3/2=1+1/2=1+1/﹙1×2﹚
√﹙1+1/2²+1/3²﹚=7/6=1+1/6=1+1/﹙2×3﹚
√﹙1+1/3²+1/4²﹚=13/12=1+1/12=1+1/﹙3×4﹚
……
∴猜想并验证√[1+1/n²+1/﹙n+1﹚²]
=√{[n²﹙n+1﹚²+﹙n+1﹚²+n²]/[n﹙n+1﹚]²}
=√{[﹙n²+n﹚²+2n²+2n+1]/[n﹙n+1﹚]²}
=√{[﹙n²+n﹚²+2﹙n²+n﹚+1]/[n﹙n+1﹚]²}
=√{[﹙n²+n+1﹚²/[n﹙n+1﹚]²}
=﹙n²+n+1﹚/[n﹙n+1﹚]
=1+1/[n﹙n+1﹚]
∴√﹙1+1/1²+1/2²﹚+√﹙1+1/2²+1/3²﹚+√﹙1+1/3²+1/4²﹚+……+√﹙1+1/2012²+1/2013²﹚
=[1+1/﹙1×2﹚]+[1+1/﹙2×3﹚]+[1+1/﹙3×4﹚]+……+[1+1/﹙2012×2013﹚]
=1×2012+[1/﹙1×2﹚+1/﹙2×3﹚+1/﹙3×4﹚+……+1/﹙2012×2013﹚]
=2012+﹙1/1-1/2+1/2-1/3+1/3-1/4+……+1/2012-1/2013﹚
=2012+﹙1-1/2013﹚
=2012+2012/2013