Question

(1-2的2次方分之1)乘以(1-3的2次方分之1)乘以(1-4的2次方分之1 ）一直乘到（1-2010的2次方分之1）计算

Answer 1

(1-1/2²)(1-1/3²)(1-1/4²)...(1-1/2010²)
=(1+1/2)(1-1/2)(1+1/3)(1-1/3)(1+1/4)(1-1/4)...(1+1/2010)(1-1/2010)
=(3/2)(1/2)(4/3)(2/3)(5/4)(3/4)...(2011/2010)(2009/2010)
=[(3/2)(4/3)(5/4)...(2011/2010)][(1/2)(2/3)(3/4)...(2009/2010)]
=[(3×4×5×...×2011)/(2×3×4×...×2010)][(1×2×3×...×2009)/(2×3×4×...×2010)]
=(2011/2)(1/2010)
=2011/4020

追问

您确定吗？

Answer 2

1/(1-2^2)=1/2 x 3/2
1/(1-3^2)=2/3 x 4/3
...
1/(1-n^2)=(n-1)/n x (n+1)/n
所有中间项将互相消去，最后剩下1/2 x 2011/2010 =2011/4020

Answer 3

等于20121212

追问

过程呢？