\(B=2+2^2+2^3+2^4+...+2^{10}\)

=>\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

=>\(B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

=>\(B=3\left(2+2^3+...+2^9\right)⋮3\left(đpcm\right)\)

Trả lời:

\(B=2+2^2+2^3+2^4+....2^9+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\) (Phần này là nhóm các lũy thừa có cùng cơ số 2 vào các nhóm sao cho tổng nhóm đầu tiên chia hết cho 3 thì mấy nhóm sau với số số hạng tương tự nhóm 1 thì oke giải tiếp như sau)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(=3\left(2+2^3+...+2^9\right)\)

Vì \(3⋮3\Rightarrow3\left(2+2^3+...+2^9\right)⋮3\)

=> đpcm

Vậy B chia hết cho 3

#Huyền Anh

\(B=2+2^2+2^3+2^4+...+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(=2\left(1+2\right)+2^2\left(1+2\right)+...+2^9\left(1+2\right)\)

\(=3\left(2+2^2+...+2^9\right)⋮3\)

\(\Rightarrow B⋮3\)

..

\(B=2+2^2+2^3+...+2^{10}\)

    =\(\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

   =\(2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

   =\(2.3+2^3.3+2^5.3+2^7.3+2^9.3\)

  =\(3\left(2+2^3+2^5+2^7+2^9\right)⋮3\)

Vậy \(B⋮3\)

\(B=2+2^2+2^3+...+2^{10}\)

\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(B=2\cdot3+2^3\cdot3+...+2^9\cdot3\)

\(B=3\left(2+2^3+...+2^9\right)\)

Vì \(3⋮3\Rightarrow3\left(2+2^3+...+2^9\right)⋮3\)

\(\Rightarrow B⋮3\)

Vậy ...

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

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

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

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

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

Vậy \(C⋮5\)

`#3107.101107`

a,

\(C=2+2^3+2^5+...+2^{23}\)

\(=\left(2+2^3+2^5\right)+\left(2^5+2^7+2^9\right)+...+\left(2^{19}+2^{21}+2^{23}\right)\)

\(=2\left(1+2^2+2^4\right)+2^5\cdot\left(1+2^2+2^4\right)+...+2^{19}\cdot\left(1+2^2+2^4\right)\)

\(=\left(1+2^2+2^4\right)\cdot\left(2+2^5+...+2^{19}\right)\)

\(=21\cdot\left(2+2^5+...+2^{19}\right)\)

Vì \(21\text{ }⋮\text{ }21\)

\(\Rightarrow21\left(2+2^5+...+2^{19}\right)\text{ }⋮\text{ }21\)

Vậy, \(C\text{ }⋮\text{ }21\)

b,

\(C=2+2^3+2^5+...+2^{23}\)

\(=\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{21}+2^{23}\right)\)

\(=\left(2+2^3\right)+2^4\cdot\left(2+2^3\right)+...+2^{20}\cdot\left(2+2^3\right)\)

\(=\left(2+2^3\right)\cdot\left(1+2^4+...+2^{20}\right)\)

\(=10\cdot\left(1+2^4+...+2^{20}\right)\)

Vì \(10\text{ }⋮\text{ }10\)

\(\Rightarrow10\cdot\left(1+2^4+...+2^{20}\right)\text{ }⋮\text{ }10\)

Vậy, \(C\text{ }⋮\text{ }10.\)

a) c = 2 + 2³ + 2⁵ + ... + 2¹⁹ + 2²¹ + 2²³

= (2 + 2³ + 2⁵) + (2⁷ + 2⁹ + 2¹¹) + ... + (2¹⁹ + 2²¹ + 2²³)

= 2.(1 + 2² + 2⁴) + 2⁷.(1 + 2² + 2⁴) + ... + 2¹⁹.(1 + 2² + 2⁴)

= 2.21 + 2⁷.21 + ... + 2¹⁹.21

= 21.(2 + 2⁷ + ... + 2¹⁹) ⋮ 21

Vậy c ⋮ 21

b) c = 2 + 2³ + 2⁵ + 2⁷ + ... + 2²¹ + 2²³

= (2 + 2³) + (2⁵ + 2⁷) + ... + (2²¹ + 2²³)

= 10 + 2⁴.(2 + 2³) + ... + 2²⁰.(2 + 2³)

= 10 + 2⁴.10 + ... + 2²⁰.10

= 10.(1 + 2⁴ + ... + 2²⁰) ⋮ 10

Vậy c ⋮ 10

NHANH NHA DNG CẦN

MA NÀO GIÚP TUI ĐI

ĐI MÀ

CÔ MÀ KIỂM TRA TUI MÀ  TUI KO LÀM Á  LÀ CÔ HẠ HÀNH KIEMR CỦA TUI Á

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\left(2+...+2^{19}\right)⋮7\)

tự làm nha

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\left(2+...+2^{19}\right)⋮7\)

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{19}\right)⋮7\)