Question

这个简便算法怎么算？？？

Answer 1

A = (1+39)*(1+39/2)*(1+39/3)*...*(1+39/50)
= (1+39)*(2+39)/2*(3+39)/3*...*(50+39)/50
= [(1+39)*(2+39)*(3+39)*...*(50+39)]/[1*2*3*...*50]
= (40*41*42*...*89)/(1*2*3*...*50)

同理：
B = (1+50)*(1+50/2)*...*(1+50/39)
= (50*52*...*89)/(1*2*...*39)

所以：
A/B = [(40*41*42*...*89)/(1*2*3*...*50)]/[(50*52*...*89)/(1*2*...*39)]
= [(40*41*42*...*89)/(50*52*...*89)]/[(1*2*3*...*50)/(1*2*...*39)]
= (40*41*42*...*49) / (40*41*42*...*50)
= 1/50

Answer 2

分子分母同时乘以1*2*3*...*50 （每个括号依次内分配一个乘数）

分子变为40*41*...*89

分母变为51*52*...*89*(40*41*...50)= 40*41*...*89

因此，原式=1