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Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{203}{3^{100}}< 3\)
\(A< \frac{3}{4}\left(đpcm\right)\)
- 1 số bài toán tương tự:
CMR: \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{100}{4^{100}}< \frac{4}{9}\)
Dạng tổng quát: CMR: \(\frac{1}{k}+\frac{2}{k^2}+\frac{3}{k^3}+\frac{4}{k^4}+...+\frac{n}{k^n}< \frac{k}{\left(k-1\right)^2}\)(k;n \(\in\) N*; k > 1)
Câu hỏi của Ngô Văn Nam - Toán lớp 6 - Học toán với OnlineMath
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3C-C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{99}}\)
\(\Rightarrow2C=1-\frac{1}{3^{99}}\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\)
\(2C=3C-C=1-\frac{1}{3^{99}}\Rightarrow C=\left(1-\frac{1}{3^{99}}\right):2=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)
A = \(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)
=> 4A = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}\)
=> 3A = \(1-\frac{1}{4^{2012}}\)
=> A = \(\frac{1-\frac{1}{4^{2012}}}{3}\)
Vậy A \(< \frac{1}{3}\)
\(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3C=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow C+3C=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4C< 1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}=D\)
Xét \(D=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(\frac{D}{3}=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(\Rightarrow D+\frac{D}{3}=1-\frac{1}{3^{100}}< 1\Rightarrow\frac{4D}{3}< 1\Rightarrow D< \frac{3}{4}\)
\(\Rightarrow4C< D< \frac{3}{4}\Rightarrow C< \frac{3}{16}\)
\(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
=> \(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
=> \(2C=1-\frac{1}{3^{99}}\)
=> \(C=\frac{1-\frac{1}{3^{99}}}{2}\)
Vì\(1-\frac{1}{3^{99}}< 1\Rightarrow\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
ngu vậy dat S roi tinh