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\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)
\(\Rightarrow2A-A=A=1-\left(\frac{1}{2}\right)^{2015}\)
Với B tương tự nhưng là lấy 3B
Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
=>\(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2017}}\)
\(A=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^3}-\frac{1}{2^3}\right)+...+\left(\frac{1}{2^{2016}}-\frac{1}{2^{2016}}\right)-\frac{1}{2^{2017}}\)
\(A=1-\frac{1}{2^{2017}}\)
Vậy: \(A=1-\frac{1}{2^{2017}}\)
Tính
a)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100}\\ \)
\(=\left(\frac{1.2.3...99}{2.3...100}\right).\left(\frac{3.4.5...101}{2.3.4...100}\right)\\ =\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
b)
\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{n^2}\\ < \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\\ \)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\\ =1-\frac{1}{n}< 1\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> 2S = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> 2S - S = ( \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\) ) - ( \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\))
S = 1 - \(\frac{1}{2^{10}}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
=> \(2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(S=1-\frac{1}{2^{10}}\)
Study well ! >_<