Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$

$=2.3+2^3.3+...+2^{99}.3$

$=3(2+2^3+...+2^{99})\vdots 3$

Ta có đpcm.

1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)\)

\(=30\left(1+2^4+...+2^{96}\right)⋮30\)

2:

\(B=3+3^2+3^3+...+3^{2022}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)

\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)

Có ai làm được không. Giúp mik với ...Thanks

Em ơi,chứng minh A gì nữa em????

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

\(A=\dfrac{1}{2\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{50\cdot50}\)

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

\(A=1\)

Vậy A=1

Lời giải:
Đặt $A=1+2^2+2^4+....+2^{100}$

$A=(1+2^2+2^4)+(2^6+2^8+2^{10})+.....+(2^{96}+2^{98}+2^{100})$

$A=(1+2^2+2^4)+2^6(1+2^2+2^4)+....+2^{96}(1+2^2+2^4)$

$=(1+2^2+2^4)(1+2^6+....+2^{96})$

$=21(1+2^6+....+2^{96})\vdots 21$ 

Ta có đpcm.