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`

B>A?

Tham khảo:

https://lazi.vn/edu/exercise/so-sanh-a-1-2-3-4-5-6-99-100-va-b-1-10

\(B>A\)

1/3²< 1/2.3

1/4²< 1/3.4

...

1/100²< 1/99.100

=> 1/2² + 1/3² + 1/4² + ... + 1/100²< 1/2² + 1/2.3 + 1/3.4 + ... + 1/99.100

A < 1/4 + 1/2 -1/3 + 1/3 - 1/4 +... + 1/99 - 1/100

A < 1/4 + 1/2 -1/100 < 1/4 + 1/2 = 3/4

=> A < 3/4

\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)

\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)

\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)

\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)

Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B

Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2022}}$

$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2021}}$

$\Rightarrow 3A-A=1-\frac{1}{3^{2022}}$

$\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{2022}}$

Xét hiệu:
$A-B=\frac{1}{2}-\frac{1}{2.3^{2022}}-(1-\frac{1}{3^{2021}})$

$=\frac{1}{3^{2021}}-\frac{1}{2.3^{2022}}-\frac{1}{2}$

$=\frac{5}{2.3^{2022}}-\frac{1}{2}$

$< \frac{1}{2}-\frac{1}{2}=0$

$\Rightarrow A< B$

`A = 1/3 +1/3^2 +1/3^3 +...+1/3^2022`

`<=> 3A = 1 +1/3 +1/3^2 +...+ 1/3^2021`

`=>2A =3A-A =1+1/3 +1/3^2 +..+ 1/3^2021 - 1/3-1/3^2-1/3^3..-1/3^2022`

`2A = 1-1/3^2022`

`=> A = (1-1/3^2022) :2`

Ta thấy `1- 1/3^2022 < 1-1/3^2021`

`=> (1 -1/3^2022):2<1-1/3^2021`

Hay `A<B`

\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)

\(A=\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)

\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)

\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)

\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)

\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)

mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)

\(\Rightarrow A< B\)

       A =      \(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+............+\(\dfrac{1}{3^{2023}}\)

     3A = 1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2022}}\)

3A - A =  1 - \(\dfrac{1}{3^{2023}}\)

   2A   = 1 - \(\dfrac{1}{3^{2023}}\) < 1

      B =  \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{12}\)

      B  = \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\)

     B   = \(\dfrac{8}{12}\)

     B   = \(\dfrac{2}{3}\) ⇒ 2B = \(\dfrac{4}{3}\) > 1 

2A < 2B ⇒ A < B 

\(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)

=>2S=1-1/3^100

=>S=1/2-1/2*3^100<1/2

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 ) )