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a, Ta có :\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\\ \Rightarrow2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\\ \Rightarrow2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{50}}\right)\\ \Rightarrow A=1-\dfrac{1}{2^{50}}< 1\\ \Rightarrow A< 1\) Vậy \(A< 1\)
b, Ta có :
\(B=\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}\)Vậy \(B< \dfrac{1}{2}\)
c, Ta có :
\(C=\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\\ \Rightarrow4C=1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\\\Rightarrow4C-C=\left(1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\right)-\left(\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\right)\\ \Rightarrow3C=1-\dfrac{1}{4^{1000}}< 1\\ \Rightarrow C< \dfrac{1}{3}\)Vậy \(C< \dfrac{1}{3}\)
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=1-1+\dfrac{1}{72}\)
\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)
\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)
\(=-\left(-\dfrac{173}{1287}\right)\)
\(=\dfrac{173}{1287}\)
c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{-49}{50}\)
a, \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ 3B=3+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\\ 3B-B=\left(3+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\\2B=3-\dfrac{1}{3^{2005}}\\ B=\dfrac{3-\dfrac{1}{3^{2005}}}{2}\)
b,
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\\ 5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\\ 5A-A=\left(5+5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}\right)\\ 4A=5^{51}-1\\ A=\dfrac{5^{51}-1}{4}\)
c,
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2-1}\right)......\left(\dfrac{1}{100^2-1}\right)\\ A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)......\left(\dfrac{1}{10000}-1\right)\\ A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\cdot\cdot\cdot\dfrac{9999}{10000}\\ A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\cdot\cdot\cdot\dfrac{99\cdot101}{100\cdot100}\\ A=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\\ A=\dfrac{1}{100}\cdot\dfrac{101}{2}\\ A=\dfrac{101}{200}\)
d,
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\\ A=\left(2^{100}+2^{98}+...+2^2\right)-\left(2^{99}+2^{97}+...+2^1\right)\)
Đặt \(A=B-C\)
\(\Rightarrow B=\left(2^{100}+2^{98}+...+2^2\right)vàC=\left(2^{99}+2^{97}+...+2^1\right)\)
\(B=2^{100}+2^{98}+...+2^2\\ 4B=2^{102}+2^{100}+...+2^4\\ 4B-B=\left(2^{102}+2^{100}+...+2^4\right)-\left(2^{100}+2^{98}+...+2^2\right)\\ 3B=2^{102}-2^2\\ B=\dfrac{2^{102}-2^2}{3}\left(1\right)\)
\(C=2^{99}+2^{97}+...+2^1\\ 4C=2^{101}+2^{99}+...+2^3\\ 4C-C=\left(2^{101}+2^{99}+...+2^3\right)-\left(2^{99}+2^{97}+...+2\right)\\ 3C=2^{101}-2\\ C=\dfrac{2^{101}-2}{3}\left(2\right)\)
Từ (1) và (2) ta có :
\(A=\dfrac{2^{102}-2^2}{3}-\dfrac{2^{101}-2}{3}\\ A=\dfrac{2^{102}-2^2-2^{101}+2}{3}\\ A=\dfrac{2^{102}-2^{101}+2}{3}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(A=2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)\(\Rightarrow A=1-\dfrac{1}{2^{100}}\)
a) \(\dfrac{-5}{9}.\dfrac{3}{11}+\dfrac{-13}{18}.\dfrac{3}{11}\)
\(=\dfrac{3}{11}.\left(\dfrac{-5}{9}+\dfrac{-13}{9}\right)\)
\(=\dfrac{3}{11}.\left(-2\right)\)
\(=\dfrac{-6}{11}\)
b) \(\dfrac{11}{2}.2\dfrac{1}{3}-1\dfrac{1}{5}.1\dfrac{1}{2}\)
\(=\dfrac{11}{3}.\dfrac{7}{3}-\dfrac{6}{5}.\dfrac{3}{2}\)
\(=\dfrac{77}{9}-\dfrac{9}{5}\)
\(=\dfrac{385}{45}-\dfrac{81}{45}\)
\(=\dfrac{304}{45}\)
c) \(1\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}-\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{10}{9}.\dfrac{2}{145}-\dfrac{8}{3}\)
\(=\dfrac{4}{261}-\dfrac{8}{3}\)
\(=\dfrac{4}{261}-\dfrac{696}{261}\)
\(=-\dfrac{692}{261}\)
d) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)+\left(2-2\right)+\left(3-3\right)+4-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=0+0+0+4-1-1-1\)
\(=4-3\)
\(=1\)
T làm biếng lắm; làm C thôi
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)
Làm tương tự ta được A > 1/15
câu a
\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)
\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+............+\dfrac{1}{2^{100}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+.........+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)
Nguyễn Thanh Hằng Tiếp đi Hằng