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\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{2015}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)
\(2B=1+\frac{1}{2}+...+\frac{1}{2^{2014}}\)
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)
\(B=1-\frac{1}{2^{2015}}< 1\). Vậy ta có điều phải chứng minh
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
=> \(A=1-\frac{1}{2^{99}}<1\)
=> \(A<1\)(Đpcm)
Ta có:
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\)
\(2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}<1\left(đpcm\right)\)
\(A=\left(\frac{5}{1.2.3}+\frac{5.2}{2.3.4}+....+\frac{5.2014}{2014.2015.2016}\right)+\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{2014.2015.2016}\right)\)
\(A=\left(\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{2015.2016}\right)+\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)
\(A=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)+\frac{1}{2}-\frac{1}{2015}+\frac{1}{2016}\)
\(A=\frac{5}{2}-\frac{5}{2016}+\frac{1}{2}-\frac{1}{2015}+\frac{1}{2016}=3-\frac{1}{504}-\frac{1}{2015}\)
Ta có : \(\frac{1}{n^2}-1=\frac{1-n^2}{n^2}=\frac{\left(1-n\right)\left(1+1\right)}{n^2}\)
Áp dụng :
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}.....\frac{-2013.2015}{2014.2014}\)
\(=\frac{-\left(1.2.3...2013\right)\left(3.4.5....2015\right)}{\left(2.3.4.....2014\right)\left(2.3.4......2014\right)}=\frac{-2015}{2014.2}=\frac{-2015}{4028}\)
Sr còn thiếu
\(A=-\frac{2015}{4028}< \frac{-2014}{4028}=-\frac{1}{2}\)
Vậy \(A< B\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)
\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)
\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)
\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)
Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)
\(A=\frac{1}{2}+\frac{1^2}{2^2}+...+\frac{1^{2014}}{2^{2014}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{2013}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{2013}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\right)\)
\(A=1-\frac{1}{2^{2014}}< 1\)
Đpcm