\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\\ \Rightarrow3B-B=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\\ \Rightarrow2B=3^{101}-3\\ \Rightarrow B=\dfrac{3^{101}-3}{2}\)

B = 3¹ + 3² + 3³ + .... + 3⁹⁹ + 3¹⁰⁰

3B = 3(3¹ + 3² + 3³ + ..... + 3⁹⁹ + 3¹⁰⁰)

3B = 3² + 3³ + 3⁴ +...... + 3¹⁰⁰ + 3¹⁰¹

3B - B = 2B = (3² + 3³ + 3⁴ + .... + 3¹⁰⁰ + 3¹⁰¹) - ( 3¹ + 3² + 3³ + .... + 3¹⁰⁰)

2B = (3² - 3²) + (3³ - 3³) +.....+ ( 3¹⁰⁰ - 3¹⁰⁰) + ( 3¹⁰¹ - 1)

2B = 0 + 0 + 0 + ..... +0 + 3¹⁰¹ - 1

2B = 3¹⁰¹ - 1

B = (3¹⁰¹ - 1)  : 2

A = 4 + 4² + 4³ + 4⁴ + ... + 4⁹⁹ + 4¹⁰⁰

A = ( 4 + 4² ) + ( 4³ + 4⁴ ) + ... + (4⁹⁹ + 4¹⁰⁰)

A = ( 4 + 4² ) + 4³(4 + 4² ) + .... + 4⁹⁹(4 + 4²)

A = 20 + 4³.20 + .... + 4⁹⁹.20

A = 20 ( 1 + 4³ + .... + 4⁹⁹ )

A = 4.5.(1 + 4^{3 + ...} + 4⁹⁹ ) ⋮ 5 ( đpcm )

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

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

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

\(=5\left(4+4^3+...+4^{99}\right)\Rightarrow A⋮5\)

Ta có:

A = 4 + 4²  + 4³ + 4⁴ + ... + 4⁹⁹ + 4¹⁰⁰

A = (4 + 4²) + (4³ + 4⁴) + ... + (4⁹⁹ + 4¹⁰⁰)

A = 4(1 + 4) + 4³(1 + 4) + ... + 4⁹⁹(1 + 4)

A = 4.5 + 4³.5 + ... + 4⁹⁹.5

A = 5.(4 + 4³ + ... + 4⁹⁹)

Vậy A chia hết cho 5

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

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

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

\(A=4.5+4^3.5+..4^{99}.5\)

\(A=5.\left(4+4^3+...4^{99}\right)\)

\(\Rightarrow A⋮5\)

A=4+4²+4³+4⁴+......+4⁹⁹+4¹⁰⁰

=> A=(4+4²)+(4³+4⁴)+......+(4⁹⁹+4¹⁰⁰)

=> A=4(1+4)+4³(1+4)+.....+4⁹⁹(1+4)

=> A=4.5+4³.5+.....+4⁹⁹.5

=> A=5(4+4³+....+4⁹⁹)

=> A chia hết cho 5 (đpcm)

`A=4+4^2+4^3+...+4^98 +4^99`

`A=(4+4^2+4^3)+...+(4^97 +4^98 +4^99)`

`A=4(1+4+4^2)+...+4^97 (1+4+4^2)`

`A=4.21+...+4^97 .21`

`A=21.(4+4^97)  \vdots 21`

   `=>Đpcm`

$A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$

$\Rightarrow {}^{}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow {}^{}<\frac{1}{101}<\frac{1}{100}=\frac{1}{{}^{}}$

$\iff A<$

$A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$

$\Rightarrow {}^{}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow {}^{}<\frac{1}{101}<\frac{1}{100}=\frac{1}{{}^{}}$

$\iff A<$

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

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

\(A=\left(4.1+4.4\right)+\left(4^3.1+4^3.4\right)+...+\left(4^{99}.1+4^{99}.4\right)\)

\(A=4.5+4^3.5+...+4^{99}.5\)

\(A=5.\left(4+4^3+...+4^{99}\right)⋮5\left(ĐPCM\right)\)

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

\(A=4\cdot\left(1+4\right)+4^3\cdot\left(1+4\right)+...+4^{99}\cdot\left(1+4\right)\)

\(A=4\cdot5+4^3\cdot5+...+4^{99}\cdot5\)

\(A=5\cdot\left(4+4^3+...+4^{99}\right)⋮5\left(đpcm\right)\)