简便计算，急啊

1+1/[n*(n+2)]=(n+1)^2/[n*(n+2)]
原式=[2*2/(1*3)]*[3*3/(2*4)]*[4*4/(3*5)]......*[99*99/(98*100)]
列式后发现，3，4......99都会全部消去，只剩下2/100=1/50

1+1/（2n-1）*（2n+1）=(2n)^2/(2n-1)*(2n+1)(n为整数)
1+1/2n*（2n+2）=(2n+1)^2/2n*(2n+2)(n为正整数)

(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*........*(1+1/98*100)*(1+1/99*101)
=（2/1*2/3）*(3/2*3/4)*....(100/99*100/101)
=(2/1*3/2*4/3...99/98*100/99)*(2/3*3/4*...99/100*100/101)
=(100/1)*(2/101)
=200/101