\(B=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)

\(3B=1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{100}{3^{99}}\)

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

\(2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

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

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

\(4B=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)

\(4B=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)

\(4B=3-\dfrac{203}{3^{100}}< 3\)

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

nhân 3 trừ đi sau đó xét cái sau sẽ thấy B<3/4

Lười lắm

\(B=\dfrac{1}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{100}}\\ \Rightarrow3B=1+\dfrac{2}{3}+....+\dfrac{100}{3^{99}}\\ \Rightarrow2B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\\ A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^{99}}\\ \Rightarrow3A=3+1+..+\dfrac{1}{3^{98}}\\ \Rightarrow2.A=3-\dfrac{1}{3^{99}}< 3\\ \Rightarrow A< \dfrac{3}{2}\\ \Rightarrow2B=A-\dfrac{100}{3^{100}}< \dfrac{3}{2}\\ \Rightarrow B< \dfrac{3}{4}\)

Lời giải:

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

\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(A+3A=1+\frac{1-2}{3}+\frac{-2+3}{3^2}+\frac{3-4}{3^3}+\frac{-4+5}{3^4}+...+\frac{99-100}{3^{99}}-\frac{100}{3^{100}}\)

\(4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-.....+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(4A=(1-\frac{1}{3})+(\frac{1}{3^2}-\frac{1}{3^3})+...+(\frac{1}{3^{98}}-\frac{1}{3^{99}})-\frac{100}{3^{100}}\)

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

\(2A=\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{50}{3^{100}}\)

\(18A=3+\frac{1}{3}+...+\frac{1}{3^{97}}-\frac{450}{3^{100}}\)

\(\Rightarrow 18A-2A=3-\frac{1}{3^{99}}-\frac{450}{3^{100}}+\frac{50}{3^{100}}=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}\)

\(\Leftrightarrow 16A=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}<3\Rightarrow A< \frac{3}{16}\)

Đặt A=1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100

3A=1-2/3+3/3^2-4/3^3+...+99/3^98-100/3^99

3A+A=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99-100/3^100

<1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99

Đặt S=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99

3S=3-1+1/3-1/3^2+1/3^3-...-1/3^98

3S+S=3-1/3^99

S=(3-1/3^99) :4

S=3/4-1/4.3^99

\(\Rightarrow\)4A<3/4-1/4.3^99

\(\Rightarrow\)A<(3/4-1/4.3^99):4

\(\Rightarrow\)A<3/16-1/16.3^99<3/16

Vậy 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16

Đặt A = \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{100}}\)

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

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

\(\Rightarrow3B=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(2B=3B-B=3-\dfrac{1}{3^{99}}\)

Nhận xét : 2B < 3 => B < \(\dfrac{3}{2}\)

=> \(B-\dfrac{100}{3^{100}}< \dfrac{3}{2}\) hay 2A < \(\dfrac{3}{2}\)

=> Đpcm

***tik mik nhé***

Ta có :

3M=1+2/3+3/3^2+...+100/3^99

Suy ra :

2M=1+(1/3+1/3^2+1/3^3+...+1/3^99)-100/3^100

Xét B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

3B=\(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)

2B=1-\(\dfrac{1}{3^{99}}\)<1/2

Suy ra : 2M<1+1/2 nên M<3/4

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

\(6A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\\ 6A-2A=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\\ 4A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\\ A=\dfrac{3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{4}=\dfrac{3}{4}-\dfrac{\dfrac{101}{3^{99}}}{4}-\dfrac{\dfrac{100}{3^{100}}}{4}< \dfrac{3}{4}\)

Vậy ...