Question

[ ( 1 − 1 2 ^2 ) ( 1 − 1 3^2 ) . . . . . . ( 1 − 1/100^2 ) ]

Answer 1

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)

Answer 2

=(1−12)(1−13)⋅...⋅(1−1100)(1+12)(1+13)⋅...⋅(1+1100)=(1−12)(1−13)⋅...⋅(1−1100)(1+12)(1+13)⋅...⋅(1+1100)

=1100⋅1012=101200