一道算数求解

(1+2/3)*(1-2/3)*(1+2/5)*(1-2/5)*......*(1+2/99)*(1-2/99)
=(1+2/3)*(1+2/5)*......*(1+2/99)*(1-2/3)*(1-2/5)*......*(1-2/99)
=5/3*7/5*......*101/97*1/3*3/5*......*97/99
=1/3*101*1/99
=101/297

（1+2/3）*（1-2/3）*(1+2/5)*(1-2/5)*(1+2/7)*(1-2/7)*…*(1+2/99)*(1-2/99)
=[（1-2/3）*（1+2/3）]*[(1-2/5)*(1+2/5)]*[(1-2/7)*(1+2/7)]*…*[(1-2/99)*(1+2/99)]
=[(1/3)(5/3)][(3/5)(7/5)][(5/7)(9/7)]……[(97/99)(101/99)]
=(1/3)(101/99)=101/297

5/3 1/3 7/5 3/5 9/7 5/7......
把每个数字算出来就可以看出来前后分子分母可以消去
所以最后结果是1/3 1/1 101/1 1/99
101/297

(1-2/3)*[(1+2/3)*(1-2/5)....]*(1+2/99)
=(1/3)*1*(101/99)
=101/297

5/3*1/3*7/583/5*9/7*5/7……99/97*95/97*101/99*97/99 都可以约掉
最后是101/297