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a, \(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}\)
\(C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)(đpcm)
b, Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{397}{3^{100}}\)
\(A=\frac{3}{4}-\frac{397}{4.3^{100}}< \frac{3}{4}\)(đpcm)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}< 1\)
Vậy \(A< 1\left(đpcm\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}\)
\(\Leftrightarrow B< \frac{3}{4}\left(đpcm\right)\)
#)Giải :
Bài 1 :
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\Leftrightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Leftrightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(\Leftrightarrow2C=1-\frac{1}{3^{100}}\Leftrightarrow C=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\Rightarrow C< \frac{1}{2}\left(đpcm\right)\)
Bài 2 :
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{16}\right)+...+\left(\frac{1}{81}-\frac{1}{100}\right)=1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2015^2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2015}\)
\(\Rightarrow A< \approx0,75\)
Vậy.....
1) Tính C
\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{n-1}{n!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
3) a) Ta có : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)