\(B=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(B=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\dfrac{49}{1}\)

\(B=\left(\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\right)+1\)

\(B=\dfrac{50}{50}+\dfrac{50}{49}+\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}\)

\(B=50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+...+\dfrac{1}{2}\right)}=\dfrac{1}{50}\)

\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)

a/ Đặt :

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+.........+\dfrac{1}{3^{50}}\)

\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^{49}}\)

\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{50}}\right)\)

\(\Leftrightarrow2A=1-\dfrac{1}{3^{50}}\)

còn sao nx thì mk chịu =.=

2:

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

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

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

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)

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

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

2 \(\times\) B + B  =                1  -  \(\dfrac{1}{2^{100}}\)

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

             B =               ( 1 - \(\dfrac{1}{2^{100}}\)) : 3

       A              =          1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\) 

A\(\times\)  3             =   3 +  1 + \(\dfrac{1}{3}\) +  \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+  \(\dfrac{1}{3^{n-1}}\) 

A \(\times\) 3 - A        = 3 - \(\dfrac{1}{3^n}\)

       2A           = 3  - \(\dfrac{1}{3^n}\)

         A           = ( 3 - \(\dfrac{1}{3^n}\)) : 2

C = \(\dfrac{3}{2^2}\) \(\times\) \(\dfrac{8}{3^2}\) \(\times\) \(\dfrac{15}{4^2}\) \(\times\) ...........\(\times\) \(\dfrac{899}{30^2}\)

C = \(\dfrac{1\times3}{2^2}\) \(\times\) \(\dfrac{2\times4}{3^2}\) \(\times\) \(\dfrac{3\times5}{4^2}\) \(\times\)........\(\times\) \(\dfrac{29\times31}{30^2}\)

C = \(\dfrac{1\times2\times\left(3\times4\times5\times....\times29\right)^2\times30\times31}{2^2\times\left(3\times4\times5\times.......\times29\right)^2\times30^2}\)

C =  \(\dfrac{2\times\left(3\times4\times5\times.....\times29\right)^2\times30}{2\times\left(3\times4\times5\times.....\times29\right)^2\times30}\) \(\times\) \(\dfrac{1\times31}{2\times30}\)

C = 1 \(\times\) \(\dfrac{31}{60}\)

C = \(\dfrac{31}{60}\)

a, \(A=1+2+2^2+2^3+..........+2^{49}+2^{50}\)

\(\Leftrightarrow2A=2+2^2+..............+2^{50}+2^{51}\)

\(\Leftrightarrow2A-A=\left(2+2^2+.........+2^{51}\right)-\left(1+2+......+2^{50}\right)\)

\(\Leftrightarrow A=2^{51}-1\)

a) \(A=1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\)

\(\Rightarrow2A=2\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)

\(2A=2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\)

\(\Rightarrow2A-A=A=\left(2+2^2+2^3+2^4+2^5+...+2^{50}+2^{51}\right)-\left(1+2+2^2+2^3+2^4+...+2^{49}+2^{50}\right)\)

\(A=2^{51}-1\) vậy \(A=2^{51}-1\)

b) \(B=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^5+...+\left(\dfrac{1}{2}\right)^{99}+\left(\dfrac{1}{2}\right)^{100}\)

\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\)

\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)

\(2B=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+\dfrac{2}{2^4}+\dfrac{2}{2^5}+...+\dfrac{2}{2^{99}}+\dfrac{2}{2^{100}}\)

\(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)

\(\Rightarrow2B-B=B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^{99}}+\dfrac{1}{2^{100}}\right)\)

\(B=1-\dfrac{1}{2^{100}}\) vậy \(B=1-\dfrac{1}{2^{100}}\)

a) \(A=2A-A\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)

\(=1-\dfrac{1}{2^{2022}}\)

b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)

\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)

a) A = 2 A − A = 2 ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 + 1 2 + . . . + 1 2 2021 − ( 1 2 + 1 2 2 + . . . + 1 2 2022 ) = 1 − 1 2 2022 b) B = 20 + 15 + 12 + 17 60 = 4 5 = 1 − 1 5 A > B ( V ì ( 1 2 2022 < 1 5 ) )

Ta có `3A=1+1/3+....+1/3^99`

`=>3A-A=1-1/3^100`

`=>2A=1-1/3^100`

`=>A=1/2-1/(2.3^100)<1/2`

Hay `A<B`

a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

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

\(\Rightarrowđpcm\)

d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)

\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)

\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)

\(\Rightarrowđpcm\)

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)

\(\Rightarrowđpcm\)

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

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

=> 3A-A= 1-\(\dfrac{1}{3^{99}}\)

=> A=\(\dfrac{1}{2}-\dfrac{1}{3^{99}.2}\)

=> A<1/2

Vậy A<1/2