Toán Lớp 8: Kết quả của phép tính A= (1- 1/2^2)(1-1/3^2)(1-1/4^2)…(1-1/2020^2) là
A. 2019/4040
B. 2019/2020
C. 2021/4040
D. 1/2
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Toán Lớp 8: Kết quả của phép tính A= (1- 1/2^2)(1-1/3^2)(1-1/4^2)…(1-1/2020^2) là
A. 2019/4040
B. 2019/2020
C. 2021/4040
D. 1/2
Comments ( 2 )
A = \left( {1 – \dfrac{1}{{{2^2}}}} \right)\left( {1 – \dfrac{1}{{{3^2}}}} \right)…\left( {1 – \dfrac{1}{{{{2020}^2}}}} \right)\\
A = \dfrac{{{2^2} – {1^2}}}{{{2^2}}}.\dfrac{{{3^2} – {1^2}}}{{{3^2}}}…\dfrac{{{{2020}^2} – 1}}{{{{2020}^2}}}\\
A = \dfrac{{\left( {2 + 1} \right)\left( {2 – 1} \right)}}{{2.2}}.\dfrac{{\left( {3 + 1} \right)\left( {3 – 1} \right)}}{{{3^2}}}…\dfrac{{\left( {2020 – 1} \right)\left( {2020 + 1} \right)}}{{{{2020}^2}}}\\
A = \dfrac{{1.3}}{{2.2}}.\dfrac{{2.4}}{{3.3}}.\dfrac{{3.5}}{{4.4}}…\dfrac{{2019.2021}}{{2020.2020}}\\
A = \dfrac{{\left( {1.2.3…2021} \right)\left( {3.4.5…2019} \right)}}{{\left( {2.3…2020} \right)\left( {2.3…2020} \right)}}\\
A = 2021.\dfrac{1}{{2.2020}} = \dfrac{{2021}}{{4040}}\\
\to C
\end{array}$