chứng minh rẳng 1/5 + 1/5^2 + 1/5^3 +..+1/5^n < 1/4
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30 tháng 8 2020
a) Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Leftrightarrow2\cdot A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2\cdot A-A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
12 tháng 8 2015
5A = 1/5 + 2/5^2 +3/5^3 +...+ 11/5^11
=> 4A= 1/5+1/5^2 +1/5^3 +...+1/5^11 - 11/5^12
=> 20A = 1+1/5+1/5^2+...+1/5^10 - 11/5^11
=> 16A = 1-1/5^11+11/5^12-11/5^11
Vì 1-1/5^11 < 1 ; 11/5^12 -11/5^11 < 0
=> 16A < 1
=> A < 1/16
PT
3 tháng 3 2017
Bài 1:
\(\dfrac{5}{x} - \dfrac{y}{3} =\dfrac{1}{6}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{y}{3}=\dfrac{5}{x}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{2y}{6}=\dfrac{5}{x}\)
\(\Rightarrow1+\dfrac{2y}{6}=\dfrac{5}{x}\)
\(\Rightarrow x.\left(1+2y\right)=30\)
Vì \(2y\) chẵn nên \(1+2y\) lẻ
\(\Rightarrow1+2y\in\left\{\pm1;\pm3;\pm5;\pm30\right\}\)
\(\Rightarrow x\in\left\{\pm10;\pm30;\pm6;\pm2\right\}\)
PT
3 tháng 3 2017
Bài 2:
\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{\left(2n-2\right).2n}\)
\(=\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{\left(2n-2\right).2n}\right).\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)
\(=\dfrac{1}{4}-\dfrac{1}{2n.2}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(đpcm\right)\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^n}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{n-1}}\)
\(5A-A=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{n-1}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^n}\right)\)
\(4A=1-\frac{1}{5^n}< 1\)
=> \(A< \frac{1}{4}\left(đpcm\right)\)
Gọi dãy số trên là : A
Ta có : \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+......+\frac{1}{5^n}\)
\(\Rightarrow5A=1+\frac{1}{5}+\frac{1}{5^2}+......+\frac{1}{5^{n-1}}\)
\(\Rightarrow5A-A=\left(1+\frac{1}{5^2}+.....+\frac{1}{5^{n-1}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+.....+\frac{1}{5^n}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^n}< 1\)
\(\Rightarrow4A< 1\Rightarrow A< \frac{1}{4}\)