Question

[Trần Anh Khang]

Tính Y=(1-1/2^2).(1-1/3^2).(1-1/4^2).(1-1/5^2)...(1-1/2006^2).(1-1/2007^2)

Answer 1

[Đặng Văn Nhật Thịnh]

Ta có : \(Y=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{2006^2}\right)\left(1-\dfrac{1}{2007^2}\right)\)

\(\Rightarrow Y=\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{2006^2-1}{2006^2}.\dfrac{2007^2-1}{2007^2}\)

\(\Rightarrow Y=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.....\dfrac{2006.2008}{2007.2007}=\dfrac{\left(1.2.....2006\right).\left(3.4.....2008\right)}{\left(2.3.....2007\right)\left(2.3.....2007\right)}=\dfrac{1.2018}{2007.2}=\dfrac{2018}{4014}\)

Answer 2

[Đoàn Đức Hà]

\(Y=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2007^2}\right)\)

\(=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.....\dfrac{2007^2-1}{2007^2}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.....\dfrac{2006.2008}{2007^2}\)

\(=\dfrac{\left(1.3\right).\left(2.4\right).....\left(2006.2008\right)}{2^2.3^2.....2007^2}\)

\(=\dfrac{\left(1.2.3.....2006\right).\left(3.4.5.....2008\right)}{\left(2.3.4.....2007\right).\left(2.3.4.....2007\right)}\)

\(=\dfrac{1.2008}{2007.2}=\dfrac{1004}{2007}\)