A = 1/2 ² + 1/3 ² + 1/4 ² +…+ 1/9 ². Chứng tỏ 8/9 > A > 2/5. Làm giúp mình cho 5 sao 20/09/2021 Bởi Eloise A = 1/2 ² + 1/3 ² + 1/4 ² +…+ 1/9 ². Chứng tỏ 8/9 > A > 2/5. Làm giúp mình cho 5 sao
Giải thích các bước giải: Ta có: $\begin{array}{l} + )A = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + … + \dfrac{1}{{{9^2}}}\\ = \dfrac{1}{{2.2}} + \dfrac{1}{{3.3}} + … + \dfrac{1}{{9.9}}\\ < \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + … + \dfrac{1}{{8.9}}\\ = \dfrac{1}{1} – \dfrac{1}{2} + \dfrac{1}{2} – \dfrac{1}{3} + … + \dfrac{1}{8} – \dfrac{1}{9}\\ = 1 – \dfrac{1}{9}\\ = \dfrac{8}{9}\\ \Rightarrow A < \dfrac{8}{9}\\ + )A = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + … + \dfrac{1}{{{9^2}}}\\ = \dfrac{1}{{2.2}} + \dfrac{1}{{3.3}} + … + \dfrac{1}{{9.9}}\\ > \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + … + \dfrac{1}{{9.10}}\\ = \dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{9} – \dfrac{1}{{10}}\\ = \dfrac{1}{2} – \dfrac{1}{{10}}\\ = \dfrac{2}{5}\\ \Rightarrow A > \dfrac{2}{5}\end{array}$ Như vậy: $\dfrac{2}{5} < A < \dfrac{8}{9}$ Bình luận
$#Leam$
$Từ$ $(1)$ $và$ $(2)$
⇒ $\dfrac{8}{9}$ > $A$ > $\dfrac{2}{5}$
CHUCBANHOKTOT ^^
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
+ )A = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + … + \dfrac{1}{{{9^2}}}\\
= \dfrac{1}{{2.2}} + \dfrac{1}{{3.3}} + … + \dfrac{1}{{9.9}}\\
< \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + … + \dfrac{1}{{8.9}}\\
= \dfrac{1}{1} – \dfrac{1}{2} + \dfrac{1}{2} – \dfrac{1}{3} + … + \dfrac{1}{8} – \dfrac{1}{9}\\
= 1 – \dfrac{1}{9}\\
= \dfrac{8}{9}\\
\Rightarrow A < \dfrac{8}{9}\\
+ )A = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + … + \dfrac{1}{{{9^2}}}\\
= \dfrac{1}{{2.2}} + \dfrac{1}{{3.3}} + … + \dfrac{1}{{9.9}}\\
> \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + … + \dfrac{1}{{9.10}}\\
= \dfrac{1}{2} – \dfrac{1}{3} + \dfrac{1}{3} – \dfrac{1}{4} + … + \dfrac{1}{9} – \dfrac{1}{{10}}\\
= \dfrac{1}{2} – \dfrac{1}{{10}}\\
= \dfrac{2}{5}\\
\Rightarrow A > \dfrac{2}{5}
\end{array}$
Như vậy: $\dfrac{2}{5} < A < \dfrac{8}{9}$