Tính giá trị của biểu thức: \(\left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right)\left( {1 - \frac{1}{{{4^2}}}} \right)...\left( {1 - \frac{1}{{{{2017}^2}}}} \right)\)
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Lời giải:
Báo sai\(\begin{array}{l}\;\;\;\left( {1 - \frac{1}{{{2^2}}}} \right)\left( {1 - \frac{1}{{{3^2}}}} \right)\left( {1 - \frac{1}{{{4^2}}}} \right)...\left( {1 - \frac{1}{{{{2017}^2}}}} \right)\\ = \frac{{\left( {{2^2} - 1} \right)\left( {{3^2} - 1} \right)\left( {{4^2} - 1} \right)...\left( {{{2017}^2} - 1} \right)}}{{{2^2}{{.3}^2}{{.4}^2}{{...2017}^2}}}\\ = \frac{{\left( {2 - 1} \right)\left( {2 + 1} \right)\left( {3 - 1} \right)\left( {3 + 1} \right)......\left( {2017 - 1} \right)\left( {2017 + 1} \right)}}{{{2^2}{{.3}^2}{{.4}^2}{{...2017}^2}}}\\ = \frac{{1.3.2.4....2016.2018}}{{{{\left( {2.3.4...2017} \right)}^2}}} = \frac{{1.2.{{\left( {3.4...2016} \right)}^2}.2017.2018}}{{{{\left( {1.2.3...2017} \right)}^2}}}\\ = \frac{{1.2.2017.2018}}{{{2^2}{{.2017}^2}}} = \frac{{2018}}{{2.2017}} = \frac{{1009}}{{2017}}.\end{array}\)
Chọn B.