C =2/3^2 +2/5^2+…+2/2019^2 cmr C <1009/2020 02/09/2021 Bởi Adalyn C =2/3^2 +2/5^2+…+2/2019^2 cmr C <1009/2020
Giải thích các bước giải: Ta có: \[{n^2} > {n^2} – 1 \Rightarrow \frac{2}{{{n^2}}} < \frac{2}{{{n^2} – 1}} = \frac{2}{{\left( {n – 1} \right)\left( {n + 1} \right)}} = \frac{{\left( {n + 1} \right) – \left( {n – 1} \right)}}{{\left( {n – 1} \right)\left( {n + 1} \right)}} = \frac{1}{{n – 1}} – \frac{1}{{n + 1}}\] Áp dụng ta có: \[\begin{array}{l}C = \frac{2}{{{3^2}}} + \frac{2}{{{5^2}}} + \frac{2}{{{7^2}}} + … + \frac{2}{{{{2019}^2}}}\\ \Rightarrow C < \frac{1}{2} – \frac{1}{4} + \frac{1}{4} – \frac{1}{6} + \frac{1}{6} – \frac{1}{8} + …. + \frac{1}{{2018}} – \frac{1}{{2020}}\\ \Rightarrow C < \frac{1}{2} – \frac{1}{{2020}} = \frac{{2009}}{{2020}}\end{array}\] Bình luận
Giải thích các bước giải:
Ta có:
\[{n^2} > {n^2} – 1 \Rightarrow \frac{2}{{{n^2}}} < \frac{2}{{{n^2} – 1}} = \frac{2}{{\left( {n – 1} \right)\left( {n + 1} \right)}} = \frac{{\left( {n + 1} \right) – \left( {n – 1} \right)}}{{\left( {n – 1} \right)\left( {n + 1} \right)}} = \frac{1}{{n – 1}} – \frac{1}{{n + 1}}\]
Áp dụng ta có:
\[\begin{array}{l}
C = \frac{2}{{{3^2}}} + \frac{2}{{{5^2}}} + \frac{2}{{{7^2}}} + … + \frac{2}{{{{2019}^2}}}\\
\Rightarrow C < \frac{1}{2} – \frac{1}{4} + \frac{1}{4} – \frac{1}{6} + \frac{1}{6} – \frac{1}{8} + …. + \frac{1}{{2018}} – \frac{1}{{2020}}\\
\Rightarrow C < \frac{1}{2} – \frac{1}{{2020}} = \frac{{2009}}{{2020}}
\end{array}\]