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NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 7 2021
\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)
Đặt \(N=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)
\(2N=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2N-N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow N=1-\dfrac{1}{2^{10}}\)
\(\Rightarrow M=1-\left(1-\dfrac{1}{2^{10}}\right)=\dfrac{1}{2^{10}}>\dfrac{1}{2^{11}}\)
Vậy \(M>\dfrac{1}{2^{11}}\)
10 tháng 7 2016
\(M=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)
\(4M=\frac{4}{4}+\frac{4}{4^2}+...+\frac{4}{4^{1000}}\)
\(4M=1+\frac{1}{4}+\frac{1}{4^2}+..+\frac{1}{4^{999}}\)
\(4M-M=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\right)\)
\(3M=1-\frac{1}{4^{1000}}\)
\(M=\left(1-\frac{1}{4^{1000}}\right):3\)
\(M=\frac{4^{1000}-1}{4^{1000}}:3\)
\(M=\frac{4^{1000}-1}{3.4^{1000}}\)
\(\frac{1}{3}=\frac{4^{1000}}{3.4^{1000}}\)
vì \(\frac{4^{1000}-1}{4^{1000}}< \frac{4^{1000}}{3.4^{1000}}\)
nên \(M< \frac{1}{3}\)
23 tháng 3 2022
\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=3\left(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+2022}\right)\)
\(=3\left(\dfrac{1}{\dfrac{2.\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3.\left(3+1\right)}{2}}+...+\dfrac{1}{\dfrac{2022.\left(2022+1\right)}{2}}\right)\)
\(=3\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2022.2023}\right)\)
\(=3.2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)\)
\(=6.\dfrac{2021}{4046}=3.\dfrac{2021}{2023}=\dfrac{6063}{2023}=\dfrac{18189}{6069}\)
\(\dfrac{10}{3}=\dfrac{20230}{6069}>\dfrac{18189}{6069}=M\)
9 tháng 1
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)
\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)
\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)
\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)
mà \(3< \dfrac{10}{3}\)
nên \(M< \dfrac{10}{3}\)
7 tháng 8 2015
\(4.M=4.\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\right)=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2014}{4^{2013}}\)
=> 4M - M = \(1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+...+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)
=> 3.M = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
Tính \(N=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)
=> \(4.N=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}\)
=> 4N - N = 4 - \(\frac{1}{4^{2013}}\)=> N = \(\frac{4}{3}-\frac{1}{3.4^{2013}}\)=> N < 4/3
Ta có: 3M < N => M < N/3 => M < (4/3)/3 = 2/9
vậy M < 4/9
DT
0
HC
0
M = 4000 > 1 = 13
M = \(1^3\)