计算:(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*(1+1/4*6)*…*(1+1/97*99)*(1+1/98*100)=?每一个括号内的通项为：1+[1/n(n+2)]=[n(n+2)+1]/[n(n+2)]=[n^2+2n+1]/[n(n+2)]=(n+1)^2/[n(n+2)]=[(n+1)/n]*[(n+1)/(n+2)]所以：原式=[(2/1)*(2/3)]*[(3/2)*(3/4)]*[(4/3)*(4/5)]*……*[(98/97)*(98/99)]*[(99/98)*(99/100)]=2*(99/100)（从第二项起，相邻两项直接约分）=99/55