Answer

(1-\dfrac{1}{2^2} )(1-\dfrac{1}{3^2})(1-\dfrac{1}{4^2}) \dots(1-\dfrac{1}{100^2})

=\dfrac{2^2-1}{2^2}\cdotp \dfrac{3^2-1}{3^2}\cdotp \dfrac{4^2-1}{4^2}\dots\dfrac{100^2-1}{100^2}

=\dfrac{1\times 3 }{2^2}\cdotp \dfrac{2\times 4 }{3^2}\cdotp \dfrac{3\times 5 }{4^2}\cdotp \dots\cdotp \dfrac{98\times 100 }{99^2}\cdotp \dfrac{99\times 101 }{100^2}

=\dfrac{1}{2}\cdotp \dfrac{101}{100}

=\dfrac{101}{200}

Steven Zheng posted 1 year ago

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