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H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
10 tháng 2 2019
\(P=\frac{1}{3^2}-\frac{1}{3^4}+....+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\)
\(\Rightarrow9P=1-\frac{1}{3^2}+....+\frac{1}{3^{2004}}-\frac{1}{3^{2006}}\)
\(\Rightarrow9P+P=\left(1-\frac{1}{3^2}+....+\frac{1}{3^{2004}}-\frac{1}{3^{2006}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+....+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\right)\)
\(\Rightarrow10P=1-\frac{1}{3^{2008}}\)
\(\Rightarrow P=\frac{1}{10}-\frac{1}{3^{2008}\cdot10}< \frac{1}{10}=0,1\)
Vậy \(P< 0,1\)
30 tháng 8 2021
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+............+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+..........+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+.........+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+.....+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+.........+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
U
0
H
0
Đặt A \(=\) \(\frac{1}{3}+\frac{2}{3^2}+...+\frac{100}{3^{100}}\)
=> 3A\(=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
=> 3A- A \(=\) 2A \(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
Đặt B \(=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)=>\(3B=3+1+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
=> 2B \(=3-\frac{1}{3^{99}}
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