cho A=1/4^2+1/6^2+1/8^2+1/10^2+…+1/160^2.Cmr1/8

28/07/2021

By 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:

$\begin{array}{l}
A = \dfrac{1}{{{4^2}}} + \dfrac{1}{{{6^2}}} + \dfrac{1}{{{8^2}}} + \dfrac{1}{{{{10}^2}}} + .. + \dfrac{1}{{{{160}^2}}}\\
 = \dfrac{1}{{{2^2}}}\left( {\dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}} \right)\\
 = \dfrac{1}{{{2^2}}}.B\\
Do:B = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}\\
 > \dfrac{1}{{{2^2}}} + \left( {\dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + … + \dfrac{1}{{80.81}}} \right)\\
 > \dfrac{1}{4} + \left( {\dfrac{1}{3} – \dfrac{1}{4} + \dfrac{1}{4} – \dfrac{1}{5} + … + \dfrac{1}{{80}} – \dfrac{1}{{81}}} \right)\\
 > \dfrac{1}{4} + \left( {\dfrac{1}{3} – \dfrac{1}{{81}}} \right)\\
 > \dfrac{1}{4} + \dfrac{{26}}{{81}} > \dfrac{1}{2}\\
 \Leftrightarrow B > \dfrac{1}{2}\\
 \Leftrightarrow A > \dfrac{1}{{{2^2}}}.\dfrac{1}{2}\\
 \Leftrightarrow A > \dfrac{1}{8}\\
Lai\,co:\\
B = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + … + \dfrac{1}{{{{80}^2}}}\\
 < \dfrac{1}{4} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + … + \dfrac{1}{{79.80}}\\
 \Leftrightarrow B < \dfrac{1}{4} + \left( {\dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{{79}} – \dfrac{1}{{80}}} \right)\\
 \Leftrightarrow B < \dfrac{1}{4} + \dfrac{1}{2} – \dfrac{1}{{80}}\\
 \Leftrightarrow B < \dfrac{3}{4} – \dfrac{1}{{80}} < \dfrac{3}{4}\\
 \Leftrightarrow A < \dfrac{1}{{{2^2}}}.B\\
 \Leftrightarrow A < \dfrac{1}{4}.\dfrac{3}{4}\\
 \Leftrightarrow A < \dfrac{3}{{16}}\\
Vậy\,\dfrac{1}{8} < A < \dfrac{3}{{16}}
\end{array}$