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Đặt \(A=\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2019^3}\)
\(\Rightarrow2A=\frac{2}{2^3}+\frac{2}{3^3}+...+\frac{2}{2019^3}\)
Ta có:
\(\left\{{}\begin{matrix}\frac{2}{2^3}< \frac{2}{1.2.3}\\\frac{2}{3^3}< \frac{1}{2.3.4}\\....\\\frac{2}{2019^3}< \frac{2}{\left(2019-1\right).2019.\left(2019+1\right)}\end{matrix}\right.\)
\(\Rightarrow2A< \frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{\left(2019-1\right).2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(2019-1\right).2019}-\frac{1}{2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2019.2020}\)
\(\Rightarrow A< \left(\frac{1}{1.2}-\frac{1}{4078380}\right):2\)
\(\Rightarrow A< \frac{1}{1.2}:2-\frac{1}{4078380}:2\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{8156760}\)
\(\Rightarrow A< \frac{1}{2^2}-\frac{1}{8156760}\)
Vì \(\frac{1}{2^2}-\frac{1}{8156760}< \frac{1}{2^2}.\)
\(\Rightarrow A< \frac{1}{2^2}\left(đpcm\right).\)
Chúc bạn học tốt!
\(=>2A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{101}}\)
\(=>2A-A=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{101}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{100}}\right)\)
\(=>A=\dfrac{1}{2^{101}}-\dfrac{1}{2}\)
nhân C vs 3 sau đó lấy 3C-C sẽ ra đc 2 C = 1 - 1/399 => C= 1/2 - 1/ (2x399 )
\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3C-C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{99}}\)
\(\Rightarrow2C=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow C=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
Mà \(1-\dfrac{1}{3^{99}}< 1\)
\(\Rightarrow C< \dfrac{1}{2}\) ( đpcm )
\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
\(3C=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
\(3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
\(3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^{99}}\right)\)\(2C=1-\dfrac{1}{3^{99}}\)
\(C=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(C=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}\)
\(C< \dfrac{1}{2}\)
\(\rightarrowđpcm\)
\(\)
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{4031}{2015^2.2016^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{2016^2-2015^2}{2015^2.2016^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)
\(A=1-\dfrac{1}{2016^2}< 1\left(đpcm\right)\)
Ta có:
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)
\(=\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+\dfrac{4}{4!}-\dfrac{1}{4!}+...+\dfrac{100}{100!}-\dfrac{1}{100!}\)
\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+...+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=1-\dfrac{1}{100!}\)
Mà \(1-\dfrac{1}{100!}< 1\)
Vậy \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) (Đpcm)
\(\dfrac{1}{2!}\)+ \(\dfrac{2}{3!}\)+ \(\dfrac{3}{4!}\)+...+\(\dfrac{99}{100!}\)
= \((\)\(\dfrac{1}{1!}\)-\(\dfrac{1}{2!}\)\()\) + \((\)\(\dfrac{1}{2!}\)-\(\dfrac{1}{3!}\)\()\) + \((\)\(\dfrac{1}{3!}\)-\(\dfrac{1}{4!}\)\()\) +...+ \((\)\(\dfrac{1}{99!}\)-\(\dfrac{1}{100!}\)\()\)
= 1-\(\dfrac{1}{100!}\) < 1.
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
\(A< \frac{9}{10}\Rightarrow A< 1\left(đpcm\right)\)
Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+......................+\dfrac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{98}}\)
\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{99}}\right)\)\(\Leftrightarrow2A=1-\dfrac{1}{3^{99}}< 1\)
\(\Leftrightarrow A< 1\)
Vậy \(\dfrac{1}{3}+\dfrac{1}{3^2}+..............+\dfrac{1}{3^{99}}< 1\rightarrowđpcm\)
Đặt:
\(S=\dfrac{1}{3}+\dfrac{1}{3^2}+.....+\dfrac{1}{3^{99}}\)
\(3S=3\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}\right)\)
\(3S=1+\dfrac{1}{3}+.....+\dfrac{1}{3^{98}}\)
\(3S-S=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}\right)\)
\(2S=1-\dfrac{1}{3^{99}}\)
\(2S< 1\)
\(S< 1\rightarrowđpcm\)