Question

[Tạ Văn Mùi]

(1-4/1)(1-4/9)(1-4/25)...(1-4/(2n-1)^2

Answer 1

[Nguyễn Văn Lâm ( ✎﹏IDΣΛ亗 )]

\(\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{4}{\left(2x+1\right)^2}\right)\) \(\left(ĐK:n\ge1\right)\)

\(=\left(\frac{1-4}{1}.\frac{9-4}{9}.\frac{25-4}{25}....\frac{\left(2n-1\right)^2-4}{\left(2n-1\right)^2}\right)\)

\(=\frac{\left(1-2\right)\left(1+2\right)}{1}.\frac{\left(3-2\right)\left(3+2\right)}{3.3}.\frac{\left(5-2\right)\left(5+2\right)}{5.5}....\frac{\left(2n-3\right)\left(2n+1\right)}{\left(2n-1\right)^2}\)

\(=\frac{\left(-1\right).3}{1}.\frac{1.5}{3.3}.\frac{3.7}{5.5}....\frac{\left(2n-3\right)\left(2n+1\right)}{\left(2n-1\right)\left(2n-1\right)}\)

\(=-\frac{2n+1}{2n-1}\)

Answer 2

[Nguyễn Văn Lâm ( ✎﹏IDΣΛ亗 )]

Sửa giúp mình dòng đầu lại thành:

\(\left(1-\frac{1}{4}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{4}{\left(2n-1\right)^2}\right)\)