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2A =2+\(\frac{3}{2^2}\)+\(\frac{4}{2^3}\)+\(\frac{5}{2^4}\)+.....+\(\frac{100}{2^{99}}\)
\(\Rightarrow\)A=2A-A=1+\(\frac{3}{4}\)\(\frac{1}{2^3}\)+\(\frac{1}{2^4}\)+.....+\(\frac{1}{2^{99}}\)-\(\frac{100}{2^{100}}\)
\(\Rightarrow\)2A=2+\(\frac{3}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+....+\(\frac{1}{2^{98}}\)-\(\frac{100}{2^{99}}\)
\(\Rightarrow\)A=2A-A=1+\(\frac{3}{4}\)+\(\frac{1}{4}\)-\(\frac{101}{2^{99}}\)+\(\frac{100}{2^{100}}\)=2-\(\frac{51}{2^{99}}\)
Cho \(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\). Chứng minh A < 2.
\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}\)
=> \(2A-A=A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Đặt \(B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}\)
=> \(2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{98}}\)
=> \(B=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}\)
=> \(A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{100}{2^{100}}-\dfrac{1}{2^{99}}\)
=> \(A=2-\dfrac{102}{2^{100}}< 2\)
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^{100}}\)
\(\dfrac{1}{2}\cdot A=\dfrac{1}{2}+\dfrac{3}{2^4}+...+\dfrac{100}{2^{101}}\)
\(A-\dfrac{A}{2}=\dfrac{1}{2A}=\dfrac{1}{2}+\dfrac{3}{2^3}+...+\dfrac{100}{2^{101}}\)
\(\left[\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right]-\dfrac{100}{2^{101}}\) (do 3/2^3=1/2^2+1/2^3)
\(\left[1-\left(\dfrac{1}{2}\right)^{101}\right]\left(1-\dfrac{1}{2}\right)-\dfrac{100}{2^{101}}\)
\(\left(\dfrac{2^{101-1}}{2^{100}}\right)-\dfrac{100}{2^{101}}\)
\(\Rightarrow A=\dfrac{\dfrac{\left(2^{101-1}\right)}{2^{99}-100}}{2^{100}}\)
Giải:
\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{3}{2^4}+\dfrac{4}{2^5}+...+\dfrac{99}{2^{100}}+\dfrac{100}{2^{101}}\)
\(A-\dfrac{A}{2}=\dfrac{1}{2A}=\dfrac{1}{2}+\dfrac{3}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}-\dfrac{100}{2^{101}}\)
\(=\left[\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right]-\dfrac{100}{2^{101}}\) ( Vì \(\dfrac{3}{2^3}=\dfrac{1}{2^2}+\dfrac{1}{2^3}\) )
\(=\dfrac{\left[1-\left(\dfrac{1}{2}\right)^{101}\right]}{\left(1-\dfrac{1}{2}\right)}-\dfrac{100}{2^{101}}\)
\(=\dfrac{\left(2^{101}-1\right)}{2^{100}}-\dfrac{100}{2^{101}}\)
\(\Rightarrow A=\dfrac{\left(2^{101}-1\right)}{2^{99}}-\dfrac{100}{2^{100}}\)
\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+.......+\dfrac{100}{2^{100}}\)
\(\Leftrightarrow2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+.........+\dfrac{100}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+......+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+.........+\dfrac{100}{2^{100}}\right)\)
\(\Leftrightarrow A=\dfrac{11}{4}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+......+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Đặt :
\(H=\dfrac{1}{2^3}+\dfrac{1}{2^4}+......+\dfrac{1}{2^{99}}\)\(\Leftrightarrow A=\dfrac{11}{4}-H-\dfrac{100}{2^{100}}\)
\(\Leftrightarrow2H=\dfrac{1}{2^2}+\dfrac{1}{2^3}+........+\dfrac{1}{2^{98}}\)
\(\Leftrightarrow2H-H=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+.....+\dfrac{1}{2^{99}}\right)\)
\(\Leftrightarrow H=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}\)
\(\Leftrightarrow A=\dfrac{11}{4}+\dfrac{1}{2^2}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)
b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)
c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)
A=1+B
B=\(\Sigma\left(\dfrac{x}{2^x}\right)\)( cho x chạy từ 3 đến 100) =1
=> A=1+B=1+1=2