cho A= 1/2^2+ 1/3^2 + …+ 1/2016^2+ 1/2017^2 chững tỏ 2/5 19/10/2021 Bởi Reese cho A= 1/2^2+ 1/3^2 + …+ 1/2016^2+ 1/2017^2 chững tỏ 2/5 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " cho A= 1/2^2+ 1/3^2 + ...+ 1/2016^2+ 1/2017^2 chững tỏ 2/5
Giải thích các bước giải: $A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}\\> \dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{2016.2017}+\dfrac{1}{2017.2018}\\= \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{2016}-\dfrac{1}{2017}+\dfrac{1}{2017}-\dfrac{1}{2018}\\= \dfrac{1}{2}-\dfrac{1}{2018}> \dfrac{1}{2} – \dfrac{1}{10}=\dfrac{2}{5}(1)\\ A = \dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{2015.2016}+\dfrac{1}{2016.2017}\\= 1 – \dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+…+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2016}-\dfrac{1}{2017}\\=1-\dfrac{1}{2017}<1(2)$ Từ (1) và (2) `=> 2/5 < A < 1 (đpcm)` Bình luận
Giải thích các bước giải:
$A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}\\> \dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{2016.2017}+\dfrac{1}{2017.2018}\\= \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{2016}-\dfrac{1}{2017}+\dfrac{1}{2017}-\dfrac{1}{2018}\\= \dfrac{1}{2}-\dfrac{1}{2018}> \dfrac{1}{2} – \dfrac{1}{10}=\dfrac{2}{5}(1)\\ A = \dfrac{1}{2^2}+\dfrac{1}{3^2}+…+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+…+\dfrac{1}{2015.2016}+\dfrac{1}{2016.2017}\\= 1 – \dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+…+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2016}-\dfrac{1}{2017}\\=1-\dfrac{1}{2017}<1(2)$
Từ (1) và (2) `=> 2/5 < A < 1 (đpcm)`