cho A=1/4^2+1/6^2+1/8^2+1/10^2+…+1/160^2.Cmr1/8 28/07/2021 Bởi Aubrey cho A=1/4^2+1/6^2+1/8^2+1/10^2+…+1/160^2.Cmr1/8 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " cho A=1/4^2+1/6^2+1/8^2+1/10^2+...+1/160^2.Cmr1/8
Đáp án: $\begin{array}{l}A = \dfrac{1}{{{4^2}}} + \dfrac{1}{{{6^2}}} + \dfrac{1}{{{8^2}}} + \dfrac{1}{{{{10}^2}}} + .. + \dfrac{1}{{{{160}^2}}}\\ = \dfrac{1}{{{2^2}}}\left( {\dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}} \right)\\ = \dfrac{1}{{{2^2}}}.B\\Do:B = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}\\ > \dfrac{1}{{{2^2}}} + \left( {\dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + … + \dfrac{1}{{80.81}}} \right)\\ > \dfrac{1}{4} + \left( {\dfrac{1}{3} – \dfrac{1}{4} + \dfrac{1}{4} – \dfrac{1}{5} + … + \dfrac{1}{{80}} – \dfrac{1}{{81}}} \right)\\ > \dfrac{1}{4} + \left( {\dfrac{1}{3} – \dfrac{1}{{81}}} \right)\\ > \dfrac{1}{4} + \dfrac{{26}}{{81}} > \dfrac{1}{2}\\ \Leftrightarrow B > \dfrac{1}{2}\\ \Leftrightarrow A > \dfrac{1}{{{2^2}}}.\dfrac{1}{2}\\ \Leftrightarrow A > \dfrac{1}{8}\\Lai\,co:\\B = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}\\ < \dfrac{1}{4} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + … + \dfrac{1}{{79.80}}\\ \Leftrightarrow B < \dfrac{1}{4} + \left( {\dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{{79}} – \dfrac{1}{{80}}} \right)\\ \Leftrightarrow B < \dfrac{1}{4} + \dfrac{1}{2} – \dfrac{1}{{80}}\\ \Leftrightarrow B < \dfrac{3}{4} – \dfrac{1}{{80}} < \dfrac{3}{4}\\ \Leftrightarrow A < \dfrac{1}{{{2^2}}}.B\\ \Leftrightarrow A < \dfrac{1}{4}.\dfrac{3}{4}\\ \Leftrightarrow A < \dfrac{3}{{16}}\\Vậy\,\dfrac{1}{8} < A < \dfrac{3}{{16}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
A = \dfrac{1}{{{4^2}}} + \dfrac{1}{{{6^2}}} + \dfrac{1}{{{8^2}}} + \dfrac{1}{{{{10}^2}}} + .. + \dfrac{1}{{{{160}^2}}}\\
= \dfrac{1}{{{2^2}}}\left( {\dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}} \right)\\
= \dfrac{1}{{{2^2}}}.B\\
Do:B = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}\\
> \dfrac{1}{{{2^2}}} + \left( {\dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + … + \dfrac{1}{{80.81}}} \right)\\
> \dfrac{1}{4} + \left( {\dfrac{1}{3} – \dfrac{1}{4} + \dfrac{1}{4} – \dfrac{1}{5} + … + \dfrac{1}{{80}} – \dfrac{1}{{81}}} \right)\\
> \dfrac{1}{4} + \left( {\dfrac{1}{3} – \dfrac{1}{{81}}} \right)\\
> \dfrac{1}{4} + \dfrac{{26}}{{81}} > \dfrac{1}{2}\\
\Leftrightarrow B > \dfrac{1}{2}\\
\Leftrightarrow A > \dfrac{1}{{{2^2}}}.\dfrac{1}{2}\\
\Leftrightarrow A > \dfrac{1}{8}\\
Lai\,co:\\
B = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}\\
< \dfrac{1}{4} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + … + \dfrac{1}{{79.80}}\\
\Leftrightarrow B < \dfrac{1}{4} + \left( {\dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{{79}} – \dfrac{1}{{80}}} \right)\\
\Leftrightarrow B < \dfrac{1}{4} + \dfrac{1}{2} – \dfrac{1}{{80}}\\
\Leftrightarrow B < \dfrac{3}{4} – \dfrac{1}{{80}} < \dfrac{3}{4}\\
\Leftrightarrow A < \dfrac{1}{{{2^2}}}.B\\
\Leftrightarrow A < \dfrac{1}{4}.\dfrac{3}{4}\\
\Leftrightarrow A < \dfrac{3}{{16}}\\
Vậy\,\dfrac{1}{8} < A < \dfrac{3}{{16}}
\end{array}$