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Câu 1.8: Giải
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(A>\dfrac{2}{5}\) (1)
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(A< 1-\dfrac{1}{9}\)
\(A< \dfrac{8}{9}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
Ta có :
\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=2009\)
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{2006.2007}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2006}-\dfrac{1}{2007}\)
\(=\dfrac{1}{4}-\dfrac{1}{2007}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4}\left(đpcm\right)\)
Vậy...
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)\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
Câu đầu sai đề nhé! Phải là 2007 chứ ko phải 20007!
\(A=\dfrac{1}{2}\cdot\dfrac{1}{7}+\dfrac{1}{7}\cdot\dfrac{1}{12}+...+\dfrac{1}{2002}\cdot\dfrac{1}{2007}\\ =\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+...+\dfrac{1}{2002\cdot2007}\\ =\dfrac{1}{5}\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+...+\dfrac{5}{2002+2007}\right)\\ =\dfrac{1}{5}\cdot\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{2002}-\dfrac{1}{2007}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{2}-\dfrac{1}{2007}\right)\\ =\dfrac{1}{5}\cdot\dfrac{2005}{4014}\\ =\dfrac{401}{4014}\)
\(B=\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)...\left(1+\dfrac{1}{2007}\right)\\B=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\cdot\cdot\dfrac{2008}{2007}\\ B=\dfrac{3\cdot4\cdot...\cdot2008}{2\cdot3\cdot...\cdot2007}\\ B=\dfrac{2008}{2}\\ B=1004 \)
\(C=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2008}\right)\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2007}{2008}\\ =\dfrac{1\cdot2\cdot...\cdot2007}{2\cdot3\cdot...\cdot2008}\\ =\dfrac{1}{2008}\)
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
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
Xét p/s A=\(\dfrac{2}{3^2}+\dfrac{2}{5^2}+...........+\dfrac{2}{2007^2}\)
A<\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...........+\dfrac{2}{2006.2008}\)
A<\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2006}-\dfrac{1}{2008}\)
A<\(\dfrac{1}{2}-\dfrac{1}{2008}\)
A<\(\dfrac{1003}{2008}\)
Ta có đpcm
Ta thấy với k \(\in\) N* thì k2 > (k - 1)(k + 1).
Thật vậy, ta có (k - 1)(k + 1) = k(k + 1) - (k + 1) = k2 + k - k - 1 = k2 - 1 < k2.
Từ đó suy ra: 32 > 2 . 4; 52 > 4 . 6; 72 > 6 . 8;...; 20072 > 2006 . 2008.
\(\Rightarrow\dfrac{2}{3^2}< \dfrac{2}{2.4};\dfrac{2}{5^2}< \dfrac{2}{4.6};\dfrac{2}{7^2}< \dfrac{2}{6.8};...;\dfrac{2}{2007^2}< \dfrac{2}{2006.2008}\)
\(\Rightarrow A< \dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2006.2008}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2006}-\dfrac{1}{2008}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2008}=\dfrac{1003}{2008}\)