hổng khó, marivan2016(mk bít nick thiệt nhưng hổng nói) làm ơn k giùm mk nha cảm ơn nhìu!!!

\(3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+....+\frac{100}{3^{99}}.\)

\(2C=3C-C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}.\)

\(2C=1+A-\frac{100}{3^{100}}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}=\frac{1}{2}\left(1-\frac{1}{3^{99}}\right)< \frac{1}{2}\)

=>\(2C=1+A-\frac{100}{3^{100}}< 1+\frac{1}{2}=\frac{3}{2}\)

\(C< \frac{3}{4}.\)

đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}\)

\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)

\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}\)

\(\Rightarrow A=\frac{3}{4}-\frac{203}{\frac{3^{100}}{4}}\le\frac{3}{4}\left(ĐPCM\right)\)

\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)

\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow4D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)

\(\Rightarrow D=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\left(đpcm\right)\)

Vậy \(D< \frac{3}{4}\)

Nguồn: @Dekisugi Hidetoshi

\(ĐặtA=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)

\(4A=3-\frac{203}{3^{100}}< 3\)

\(=>A< \frac{3}{4}\left(đpcm\right)\)

Ủng hộ mk nha ^_-

a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)

=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)

=\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)

=\(1-\frac{1}{100!}< 1\)

\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)

b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)

=\(2-\frac{1}{99}-\frac{1}{100}< 2\)

\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)

Đặt:

\(4C=3-\frac{203}{3^{100}}.\)

Mà \(3-\frac{203}{3^{100}}< 3\)

\(\Rightarrow4C< 3\)

\(\Rightarrow C< \frac{3}{4}\left(đpcm\right).\)

Vậy \(C< \frac{3}{4}.\)

Chúc bạn học tốt!