(1-1/2²)(1-1/3²)(1-1/4²)…（1-1/n²）=?这是我们试卷上的希望有心人能注意到

Question

(1-1/2²)(1-1/3²)(1-1/4²)…（1-1/n²）=?
这是我们试卷上的 希望有心人能注意到

Answer 1

解
(1-1/2²)(1-1/3²)(1-1/4²)……(1-1/n²)——平方差
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4)…………(1-1/n)(1+1/n)
=1/2×3/2×2/3×4/3×3/4×…………×(n-1)/n×(n+1)/n
=1/2×(n+1)/n
=(n+1)/(2n)