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\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)
\(\Rightarrow G=\dfrac{64}{505}\)
\(B=\dfrac{2^2}{1\cdot3}+\dfrac{3^2}{2\cdot4}+\dfrac{4^2}{3\cdot5}+...+\dfrac{99^2}{98\cdot100}\\ =\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{2\cdot4+1}{2\cdot4}+\dfrac{3\cdot5+1}{3\cdot5}+...+\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}+\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}+\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}+...+\dfrac{98\cdot100}{98\cdot100}+\dfrac{1}{98\cdot100}\\ =1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{2\cdot4}+1+\dfrac{1}{3\cdot5}+...+1+\dfrac{1}{98\cdot100}\\ =\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ =98+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ \)Gọi \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\) là A
\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{295}{198}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\dfrac{14651}{9900}=\dfrac{14651}{19800}\)
\(B=98+A=98+\dfrac{14651}{19800}=98\dfrac{14651}{19800}\)
Dễ thấy phần nguyên của B là 98
Vậy phần nguyên của B là 98
\(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}=\dfrac{5}{3}\)
`(2^2)/(1 . 3) . (3^2)/(2 . 4) . (4^2)/(3 . 5) . (5^2)/(4 . 6)`
`= 4/3 . 9/8 . 16/15 . 25/24 = 5/3`
Trước hết ta chứng minh (a-1)(a+1) + 1 = a^2 (*)
Thật vậy VT = (a-1)(a+1)+1=(a-1)a + a-1 +1 = a^2-a+a=a^2 =VP
Áp dụng (*) ta có:
\(A=\dfrac{1\cdot3+2}{2^2}+\dfrac{2\cdot4+2}{3^2}+...+\dfrac{2009\cdot2011+2}{2010^2}\\ =\dfrac{2^2+1}{2^2}+\dfrac{3^2+1}{3^2}+...+\dfrac{2010^2+1}{2010^2}=2009+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2010^2}\\ < 2009+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2009\cdot2010}\\ =2009+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2009}-\dfrac{1}{2010}=2010-\dfrac{1}{2010}< 2020< 2011\)
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{99^2}{98.100}\)
\(=\dfrac{2.2.3.3.4.4.....99.99}{1.3.2.4.3.5.....98.100}\)
\(=\dfrac{2.3.4.....99}{1.2.3.4.....98}.\dfrac{2.3.4.....99}{3.4.5.....100}\)
\(=\dfrac{99}{98}\cdot\dfrac{2}{100}\)
\(=\dfrac{99}{4900}\)