bài này nữa nè:
\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
K
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24 tháng 10 2016
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)
N
11 tháng 9 2016
ta có:
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-2\)
=\(\left(1+\frac{99}{2}+1+\frac{98}{3}+1+...+\frac{1}{100}+1\right):\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-2\)
=\(\left(\frac{101}{101}+\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\right)-2\)
=\(101\left(\frac{1}{101}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\right)-2\)
=101-2=99
