Toán ChoA=2/3mũ2 +2/5mu2+2/7mũ2+•••+2/2017mũ2 07/09/2021 By Anna ChoA=2/3mũ2 +2/5mu2+2/7mũ2+•••+2/2017mũ2
Giải thích các bước giải: $\begin{array}{l}A = \frac{2}{{{3^2}}} + \frac{2}{{{5^2}}} + \frac{2}{{{7^2}}} + … + \frac{2}{{{{2017}^2}}} = \frac{1}{{{3^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{5^2}}} + … + \frac{1}{{{{2017}^2}}} + \frac{1}{{{{2017}^2}}}\\A < \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + \frac{1}{{5.6}}… + \frac{1}{{2016.2017}} + \frac{1}{{2017.2018}}\\ = \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + \frac{1}{5} – \frac{1}{6} + … + \frac{1}{{2016}} – \frac{1}{{2017}} + \frac{1}{{2017}} – \frac{1}{{2018}}\\ = \frac{1}{2} – \frac{1}{{2018}} = \frac{{504}}{{1009}}\\Vay\,A < \frac{{504}}{{1009}}\end{array}$ Trả lời
Giải thích các bước giải:
$\begin{array}{l}
A = \frac{2}{{{3^2}}} + \frac{2}{{{5^2}}} + \frac{2}{{{7^2}}} + … + \frac{2}{{{{2017}^2}}} = \frac{1}{{{3^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{5^2}}} + … + \frac{1}{{{{2017}^2}}} + \frac{1}{{{{2017}^2}}}\\
A < \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + \frac{1}{{5.6}}… + \frac{1}{{2016.2017}} + \frac{1}{{2017.2018}}\\
= \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \frac{1}{4} – \frac{1}{5} + \frac{1}{5} – \frac{1}{6} + … + \frac{1}{{2016}} – \frac{1}{{2017}} + \frac{1}{{2017}} – \frac{1}{{2018}}\\
= \frac{1}{2} – \frac{1}{{2018}} = \frac{{504}}{{1009}}\\
Vay\,A < \frac{{504}}{{1009}}
\end{array}$