Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{9}{9}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\) (1)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{5}{10}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{4}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\) (2)
Từ (1) và (2)\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
Đặt \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2014^2}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{2015}\)
\(A>\dfrac{1}{5^2}+\dfrac{1}{6}-\dfrac{1}{150}=\dfrac{1}{5}\) (đpcm)
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}< \dfrac{1}{4}\)
\(\Rightarrowđpcm\)
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=5\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}\)
\(\Rightarrow A< \dfrac{1}{4}\)
\(A=\dfrac{\left(3+\dfrac{2}{15}+\dfrac{1}{5}\right):\dfrac{5}{2}}{\left(5+\dfrac{3}{7}-2-\dfrac{1}{4}\right):\left(4+\dfrac{43}{56}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{2}{5}}{\dfrac{89}{28}:\dfrac{267}{56}}=\dfrac{4}{3}:\dfrac{2}{3}=2\)
\(B=\dfrac{\dfrac{6}{5}:\left(\dfrac{6}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{2}{5}}=2\)
Do đó: A=B
A=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\)
5A=\(\dfrac{5}{5}+\dfrac{5}{5^2}+\dfrac{5}{5^3}+...+\dfrac{5}{5^{2014}}\)
5A=\(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}\)
5A-A=\(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\right)\)4A=\(1-\dfrac{1}{5^{2014}}\)
4A=\(\dfrac{5^{2014}-1}{5^{2014}}\)
A=\(\dfrac{5^{2014}-1}{5^{2014}}:4\)
A=\(\dfrac{5^{2014}-1}{5^{2014}}.\dfrac{1}{4}\)
\(\Rightarrow\)A<\(\dfrac{1}{4}\)
Ta có:
A = \(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\)
\(\Rightarrow\) 5A = 5\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\right)\)
\(\Rightarrow\) 5A = \(\dfrac{5}{5}+\dfrac{5}{5^2}+\dfrac{5}{5^3}+....+\dfrac{5}{5^{2014}}\)
\(\Rightarrow\) 5A = \(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{2013}}\)
\(\Rightarrow\)\(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{2013}}\right)\)-\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\right)\) = 5A - A
\(\Rightarrow\)4A= 1 - \(\dfrac{1}{5^{2014}}\)
\(\Rightarrow\) A =\(\dfrac{5^{2014}-1}{5^{2014}}\) : 4
Vậy A =\(\dfrac{5^{2014}-1}{5^{2014}}\) : 4