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\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}\)
=> \(3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{2^{2018}}-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-...-\frac{2019}{4^{2019}}\)
=>3S=\(1+\frac{1}{4}+\frac{1}{4^2}+..+\frac{1}{2^{2018}}-\frac{2019}{4^{2019}}\)
còn lại tự giải nhé
\(5A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
\(\Rightarrow4A=5A-A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt \(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
Khi đó \(4A=B-\frac{99}{5^{100}}< B\)
\(5B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}\)
\(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}+\frac{1}{5^{99}}\)
\(\Rightarrow4B=5B-B=1-\frac{1}{5^{99}}\)
\(\Rightarrow B=\frac{1}{4}-\frac{1}{4\cdot5^{99}}< \frac{1}{4}\)
\(\Rightarrow4A < B\Rightarrow4A< \frac{1}{4}\)
\(\Rightarrow A< \frac{1}{16}\) ( đpcm )
2. \(M=\left(1+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(M=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)
\(\Rightarrow\left(M-N\right)^3=0\)
Xét \(4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+\dfrac{4}{4^3}+...+\dfrac{2014}{4^{2013}}\)
=> \(3S=4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2014}{4^{2013}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+...+\dfrac{2014}{4^{2014}}\right)\)
=> \(3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}-\dfrac{2014}{4^{2014}}< 1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
Đặt \(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\)
=> \(4A=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\)
=> \(3A=4A-A=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{2012}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2013}}\right)\)
=> \(3A=4-\dfrac{1}{4^{2013}}< 4\)
=> \(A< \dfrac{4}{3}\)
=> \(3S< \dfrac{4}{3}\)
=> \(S< \dfrac{4}{9}< \dfrac{1}{2}\)
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+....+\frac{2014}{4^{2013}}-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+......+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
đặt \(A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2023}}\)
\(4A-A=4+1+\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{2022}}-\left(1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2023}}\right)\)
\(3A=4-\frac{1}{4^{2023}}\)
\(A=\frac{4}{3}-\frac{1}{3.4^{2023}}\)
\(\Rightarrow3S=\frac{4}{3}-\frac{1}{3.4^{2023}}-\frac{2014}{4^{2024}}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}\)
do \(\frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow S=\frac{4}{9}-\frac{1}{9.4^{2023}}-\frac{2014}{3.4^{2024}}< \frac{4}{8}=\frac{1}{2}\left(đpcm\right)\)
Ta có:
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}=\frac{1}{2020}\)
Theo bài ra ta có:
\(\frac{1+7A}{1+9A}=\frac{1+7\cdot\frac{1}{2020}}{1+9\cdot\frac{1}{2020}}=\frac{9\left(1+7\cdot\frac{1}{1010}\right)}{7\left(1+9\cdot\frac{1}{1010}\right)}=\frac{9}{7}\)
\(=>\frac{1+7A}{1+9A}\)là phân số tối giản (ĐPCM)
Bạn giải sai rồi
Cái chỗ \(\frac{9\left(1+7.\frac{1}{1010}\right)}{7\left(1+9.\frac{1}{1010}\right)}\) ở trong ngoặc có số 7 và 9 không giống nhau nên không thể rút gọn
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}.\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}-\frac{1}{4}-\frac{2}{4^2}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(3S< A=1+\frac{1}{4}+...+\frac{1}{4^{2018}}\)\(\Rightarrow3A=4A-A=4-\frac{1}{4^{2018}}< 4\)(sau khi rút gọn)
\(\Rightarrow3.3S< 4\Rightarrow9S< 4\)
\(\Rightarrow S< \frac{4}{9}< \frac{1}{2}\)