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\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{2016}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2016}}\)
\(2A=2+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\)
\(2A-A=2+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{2015}}-\frac{1}{2}-\frac{1}{2^2}-..-\frac{1}{2^{2016}}\)
\(A=2-\frac{1}{2^{2016}}\)
\(A=\frac{2^{2009}}{2^{2010}}-\frac{1}{2^{2010}}=\frac{2^{2009}-1}{2^{2010}}\)
\(\Rightarrow A< 1\)
không biết khó quá mà bạn biết bài này không giúp mình với mình cần gấp nha nick mình là Quách Ngọc Minh Xuân
A= \(1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)
B=\(\left(\frac{2012}{1}-1\right)+\left(\frac{2012}{2}-1\right)+...+\left(\frac{2012}{2011}-1\right)\)
= \(\frac{2012}{1}-\frac{2012}{2012}+\frac{2012}{2}-\frac{2012}{2012}+...+\frac{2012}{2011}-\frac{2012}{2012}\)
=\(2012\left(1-\frac{1}{2012}+\frac{1}{2}-\frac{1}{2012}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}\)=\(\frac{2012\left(1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}{1-\frac{2011}{2012}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)= 2012
\(A< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{49.50.51}.\)
\(2A< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{49.50.51}\)
\(2A< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{51-49}{49.50.51}\)
\(2A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{49.50}-\frac{1}{50.51}\)
\(2A< \frac{1}{2}-\frac{1}{50.51}< \frac{1}{2}\Rightarrow A< \frac{1}{4}< \frac{1}{2}\)
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\)
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\left(1\right)\)
Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}< 1\left(2\right)\)
Từ (1) và (2) \(\Rightarrow A< B< 1\Rightarrow A< 1\)
Đpcm