m = 1-(1/2^2+1/3^2+1/4^2+1/5^2+...+1/10^2) . so sánh m với 0
K
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19 tháng 3 2017
M = \(\frac{1}{2}\)+ \(\frac{2}{3}\)+ \(\frac{3}{4}\)+\(\frac{4}{5}\)+ \(\frac{5}{6}\)+ \(\frac{6}{7}\)+ \(\frac{7}{8}\)+ \(\frac{8}{9}\)+ \(\frac{9}{10}\)= \(\frac{17819}{2520}\)
Vậy: M > 1
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 \(M>\dfrac{1}{2^{11}}\)
TT
0
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 \(M< \dfrac{10}{3}\)
NP
0