Question

Tính

P = (1 - \(\dfrac{1}{1^2}\)) . (1 - \(\dfrac{1}{3^2}\)) ... (1 - \(\dfrac{1}{100^2}\)

Answer 1

\(P=\left(1-\dfrac{1}{1^2}\right)\cdot\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right)\\ =0\cdot\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right)\\ =0\)

Answer 2

Ta có:

P = (1 - \(\dfrac{1}{1^2}\)).(\(1-\dfrac{1}{3^2}\))...(1-\(\dfrac{1}{100^2}\))

P = 0 . (\(1-\dfrac{1}{3^2}\))...(1-\(\dfrac{1}{100^2}\))

P = 0

Vậy P = 0