Question

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P=(1-1/2^2)(1-1/3^2)..... (1-1/100^2)

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Answer 1

\(P=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)......\left(\frac{100^2-1}{100^2}\right)=\frac{1.3}{2.2}.\frac{2.4}{3.3}....\frac{99.101}{100.100}=\frac{\left(1.2.3....99\right)}{2.3....100}.\frac{3.4.....101}{2.3.....100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)