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\(\frac{C}{3}=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)
\(\frac{2C}{3}=C-\frac{C}{3}=\frac{1}{3}-\frac{1}{3^{100}}\)
\(2C=1-\frac{1}{3^{99}}\Rightarrow C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)
C=.................................
\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Rightarrow2C=1-\frac{1}{3^{99}}\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}\)
Dễ thấy \(1-\frac{1}{3^{99}}< 1\Leftrightarrow\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\Leftrightarrow C< \frac{1}{2}\)
b) Đặt \(C=\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{1000}}\)
\(\frac{1}{4}A=\frac{1}{4^2}+\frac{1}{4^3}+.......+\frac{1}{4^{1001}}\)
\(A-\frac{1}{4}A=\left(\frac{1}{4^2}-\frac{1}{4^2}\right)+\left(\frac{1}{4^3}-\frac{1}{4^3}\right)+.....+\frac{1}{4}-\frac{1}{4^{1001}}\)
\(\frac{3}{4}A=\frac{1}{4}-\frac{1}{4^{1001}}\)
Đến đây Đặt \(\frac{3}{4}B=\frac{1}{4}\)
Ta có: \(\frac{3}{4}A<\frac{3}{4}B\) \(\rightarrow A
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
\(3A=3\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A-A=2A\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{1}{3^1}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(=1-\frac{1}{3^{100}}\)
\(2A=1-\frac{1}{3^{100}}\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
C=\(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
3C=3.( \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\) )
3C-C=( \(1+\frac{1}{3}+...+\frac{1}{3^{98}}\) ) - ( \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\) )
2C= 1 - \(\frac{1}{3^{99}}\)< 1
\(\Rightarrow\)C= \(\left(1-\frac{1}{3^{99}}\right)\div2\)<\(\frac{1}{2}\)
Điều Phải Chứng Minh