Cho A= 13/25+9/10-11/15+13/21-15/28+17/36-…+197/4851-199/4950. chứng minh A>9/10 28/07/2021 Bởi Amaya Cho A= 13/25+9/10-11/15+13/21-15/28+17/36-…+197/4851-199/4950. chứng minh A>9/10
Đáp án: $\begin{array}{l}A = \dfrac{{13}}{{25}} + \dfrac{9}{{10}} – \dfrac{{11}}{{15}} + \dfrac{{13}}{{21}} – \dfrac{{15}}{{28}} + \dfrac{{17}}{{26}} – … + \dfrac{{197}}{{4851}} – \dfrac{{199}}{{4950}}\\ = \dfrac{{13}}{{25}} + \left( {\dfrac{9}{{10}} – \dfrac{{11}}{{15}}} \right) + \left( {\dfrac{{13}}{{21}} – \dfrac{{15}}{{28}}} \right) + … + \left( {\dfrac{{197}}{{4851}} – \dfrac{{199}}{{4950}}} \right)\\ = \dfrac{{13}}{{25}} + \dfrac{1}{6} + \dfrac{1}{{12}} + … + \dfrac{1}{{2450}}\\ = \dfrac{{13}}{{25}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + … + \dfrac{1}{{49.50}}\\ = \dfrac{{13}}{{25}} + \dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{{49}} – \dfrac{1}{{50}}\\ = \dfrac{{13}}{{25}} + \dfrac{1}{2} – \dfrac{1}{{50}}\\ = 1 > \dfrac{9}{{10}}\\Vậy\,A > \dfrac{9}{{10}}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
A = \dfrac{{13}}{{25}} + \dfrac{9}{{10}} – \dfrac{{11}}{{15}} + \dfrac{{13}}{{21}} – \dfrac{{15}}{{28}} + \dfrac{{17}}{{26}} – … + \dfrac{{197}}{{4851}} – \dfrac{{199}}{{4950}}\\
= \dfrac{{13}}{{25}} + \left( {\dfrac{9}{{10}} – \dfrac{{11}}{{15}}} \right) + \left( {\dfrac{{13}}{{21}} – \dfrac{{15}}{{28}}} \right) + … + \left( {\dfrac{{197}}{{4851}} – \dfrac{{199}}{{4950}}} \right)\\
= \dfrac{{13}}{{25}} + \dfrac{1}{6} + \dfrac{1}{{12}} + … + \dfrac{1}{{2450}}\\
= \dfrac{{13}}{{25}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + … + \dfrac{1}{{49.50}}\\
= \dfrac{{13}}{{25}} + \dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{{49}} – \dfrac{1}{{50}}\\
= \dfrac{{13}}{{25}} + \dfrac{1}{2} – \dfrac{1}{{50}}\\
= 1 > \dfrac{9}{{10}}\\
Vậy\,A > \dfrac{9}{{10}}
\end{array}$