Cho biết
\(P=\dfrac{1}{2^0}+\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{1992}{2^{1991}}\)
chứng minh P<4
K
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HN
22 tháng 11 2017
Ta có: \(2P=2+\dfrac{2}{2^0}+\dfrac{3}{2^1}+...+\dfrac{1992}{2^{1990}}\)
\(\Rightarrow2P-P=\left(2+\dfrac{2}{2^0}+\dfrac{3}{2^1}+...+\dfrac{1992}{2^{1990}}\right)-\left(\dfrac{1}{2^0}+\dfrac{2}{2^1}+...+\dfrac{1992}{2^{1991}}\right)\)
\(=2-\dfrac{1992}{2^{1991}}+\left(\dfrac{1}{2^0}+\dfrac{1}{2^1}+...+\dfrac{1}{2^{1990}}\right)\)
\(=2-\dfrac{1992}{2^{1991}}+2-\dfrac{1}{2^{1990}}< 4\)
13 tháng 5 2017
Ta có :
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+......................+\dfrac{1}{50^2}\)
Ta thấy :
\(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
............................
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....................+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}\)
\(\Rightarrow A< 2-\dfrac{1}{50}< 2\)
\(\Rightarrow A< 2\rightarrowđpcm\)
8 tháng 4 2017
Ta có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{2^2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{9}=\dfrac{23}{36}< \dfrac{32}{36}=\dfrac{8}{9}\). (1)
Ta lại có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{19}{20}>\dfrac{8}{20}=\dfrac{2}{5}\). (2)
Từ (1) và (2) suy ra đpcm.
25 tháng 3 2017
Ta có:
A=\(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
A<\(1+\dfrac{1}{2.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
A<\(1+\dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
A<\(\dfrac{5}{4}\)+\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{99}-\dfrac{1}{100}\)
A<\(\dfrac{5}{4}+\dfrac{1}{2}-\dfrac{1}{100}\)
A<\(\dfrac{5}{4}+\dfrac{49}{100}\)
A<\(\dfrac{174}{100}\)<\(\dfrac{7}{4}\)
=>A<\(\dfrac{7}{4}\)
Tick giùm mink nha :D
MN
1
MV
5 tháng 5 2017
\(A=\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{2016}{2^{2017}}\\ 2A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2016}{2^{2016}}\\ 2A-A=\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2016}{2^{2016}}\right)-\left(\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{2016}{2^{2017}}\right)\\ A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{2016}{2^{2017}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2016}{2^{2016}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2016}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{2016}{2^{2017}}\right)\\ A=1-\dfrac{2017}{2^{2016}}-\dfrac{2016}{2^{2017}}\\ A=1-\dfrac{4034}{2^{2017}}-\dfrac{2016}{2^{2017}}\\ A=1-\left(\dfrac{4034}{2^{2017}}+\dfrac{2016}{2^{2017}}\right)\\ A=1-\dfrac{6050}{2^{2017}}< 1\)
Vậy \(A< 1\)
28 tháng 10 2017
Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow2^2A=1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{98}}\)
\(\Rightarrow2^2A-A=\left(1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\right)\)
\(\Rightarrow3A=1-\dfrac{1}{2^{100}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{2^{100}}}{3}< \dfrac{1}{3}\)(đpcm)
29 tháng 4 2017
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
30 tháng 4 2017
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
27 tháng 4 2017
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{50^2-1}\)
\(\Leftrightarrow A< 1+\dfrac{1}{3}+\dfrac{1}{8}+...+\dfrac{1}{2499}\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+...++\dfrac{1}{48}-\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{51}+\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)\)
Nhận xét \(\dfrac{50}{51}< 1;\dfrac{24}{50}< 1\Rightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)< 1+\dfrac{1}{2}\cdot\left(1+1\right)=2\)
Vậy A<2
BN
27 tháng 4 2017
Nhận xét: \(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...........
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
Vậy A < 2