趣味的数学-137
求
[1- (1/4)]*[1- (1/9)]*[1- (1/16)]...*[1- (1/9801)]*[1- (1/10000)]
【改编自AHSME】
对
101/200
[1- (1/4)]*[1- (1/9)]*[1- (1/16)]...*[1- (1/9801)]*[1- (1/10000)] =
[1- (1/2^2)]*[1- (1/3^2)]*[1- (1/4^2)]...*[1- (1/99^2)]*[1- (1/100^2)] =
[1*3/2^2]*[2*4/3^3]*[3*5/4^4]...*[98*100/99^2)]*[99*101/100^2] =
1/2 * 101/100 =
