\(E=25\left[3\cdot\left(5+4^2+4^3+...+4^{2021}\right)+1\right]\)

\(=25\cdot\left(4^2+4^2+4^3+...+4^{2021}\right)\)

\(=25\cdot4^{2022}⋮4^{2022}\)

Lời giải:
Đặt $A=1-2+2^2-2^3+2^4-2^5+2^6-....-2^{2021}+2^{2022}$

$A=1+(-2+2^2-2^3)+(2^4-2^5+2^6)+(-2^7+2^8-2^9)+...+(2^{2020}-2^{2021}+2^{2022})$

$A=1+(-2+2^2-2^3)+2^3(2-2^2+2^3)+2^6(-2+2^2-2^3)+....+2^{2019}(2-2^2+2^3)$

$=1+(-6)+2^3.6+2^6(-6)+....+2^{2019}.6$

$=1+6(-1+2^3-2^6+...+2^{2019})$

Suy ra $A$ chia $6$ dư $1$/

2 x 26 x 5 =............... không biết 

ta có:

tui lớp 8 ko bt làm :)

trời ơi cíu tui

\(A=8\left(1+8\right)+8^3\left(1+8\right)+...+8^{2021}\left(1+8\right)\)

\(=8.9+8^3.9+...+8^{2021}.9=9\left(8+8^3+...+8^{2021}\right)⋮9\)

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

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

\(=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{2021}.\left(1+4\right)\)

\(=4.5+4^3.5+...+4^{2021}.5\)

\(=5.\left(4+4^3+...+4^{2021}\right)⋮5\)

Vậy \(C⋮5\)

Đặt A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²²

= (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2²⁰²⁰ + 2²⁰²¹ + 2²⁰²²)

= 2.(1 + 2 + 2²) + 2⁴.(1 + 2 + 2²) + ... + 2²⁰²⁰.(1 + 2 + 2²)

= 2.7 + 2⁴.7 + ... + 2²⁰²⁰.7

= 7.(2 + 2⁴ + ... + 2²⁰²⁰) ⋮ 7

Vậy A ⋮ 7