\(3+3^2+3^3+...+3^{2012}\)

\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)

\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)

\(=40\left(3+...+3^{2009}\right)⋮40\)

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rrrrr

Ta có : A=2²+2⁴+2⁶+...+2²⁰

=(2²+2⁴)+(2⁶+2⁸)+...+(2¹⁸+2²⁰)

=2²(1+2²)+2⁶(1+2²)+...+2¹⁸(1+2²)

=2².5+2⁶.5+...+2¹⁸.5 chia hết cho 5

Vậy A chia hết cho 5.

\(A=2^2+2^4+2^6..+2^{18}+2^{20}\)

\(\Leftrightarrow A=\left(2^2+2^4\right)+\left(2^6+2^8\right)+...+\left(2^{18}+2^{20}\right)\)

\(\Leftrightarrow A=20+2^4.\left(2^2+2^4\right)+...+2^{16}.\left(2^2+2^4\right)\)

\(\Leftrightarrow A=20+2^4.20+..+2^{16}.20\)

\(\Leftrightarrow A=20\left(1+2^4+..+2^{16}\right)\)

Vì \(20⋮5\)

\(\Rightarrow A=20\left(1+2^4+..+2^{16}\right)⋮5\)

Vậy \(A⋮5\)

hok tốt!!

\(A=2^2+2^4+2^6+2^8+....+2^{18}+2^{20}\)

\(A=\left(2^2+2^4\right)+\left(2^6+2^8\right)+....+\left(2^{18}+2^{20}\right)\)

\(A=2^2\left(1+2^2\right)+2^6\left(1+2^2\right)+....+2^{18}\left(1+2^2\right)\)

\(A=2^2\cdot5+2^6\cdot5+...+2^{18}\cdot5\)

\(A=5\left(2^2+2^6+...+2^{18}\right)\)

=> A chia hết cho 5

\(A=\left(1+3+3^2\right)+...+\left(3^{99}+3^{100}+3^{101}\right)\\ A=\left(1+3+3^2\right)+...+3^{99}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(1+...+3^{99}\right)=13\left(1+...+3^{99}\right)⋮13\)

a.    A= 2+2²+2³+......+2⁶⁰

= 2+ (2²+2³)+(2⁴+2⁵)+......+(2⁵⁸+2⁵⁹)+2⁶⁰

=2+2(2+2²)+2³(2+2²)+......+2⁵⁷(2+2²)+2⁶⁰

=2+(2+2²)(2+2³......+2⁵⁷)+2⁶⁰

=2+ 6(2+2^3+......+2^57)+2⁶⁰ => cả 23 số hạng đều chia hết cho 2 => tổng chia hết cho 2 => a chia hết cho 2

b. A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+.........+(2^57+2^58+2^59+2^60)

=2(1+2+2^2+2^3)+2^5(1+2+2^2+2^3)+......+2^57(1+2+2^2+2^3)

=2.15 +2^5.15+...........+2^57.15 = 15 (2+2^5+...........+2^57) => 15 chia hết cho 3 => A chia hết cho 3

k đúng cho mình nha!!!!

a. Do 2; 2²; 2³; ...; 2⁶⁰ chia hết cho 2

=> A chia hết cho 2 ( đpcm)

b. A = 2 + 2² + 2³ + ... + 2⁶⁰ ( có 60 số; 60 chia hết cho 2)

A = (2 + 2²) + (2³ + 2⁴) + ... + (2⁵⁹ + 2⁶⁰)

A = 2.(1 + 2) + 2³.(1 + 2) + ... + 2⁵⁹.(1 + 2)

A = 2.3 + 2³.3 + ... + 2⁵⁹.3

A = 3.(2 + 2³ + ... + 2⁵⁹) chia hết cho 3

=> A chia hết cho 3 ( đpcm)

A=2+(2²+2³⁾+(2⁴+2⁵)+...+(2⁵⁹⁺2⁶⁰)

  =2+2.2²+2.2⁴+...+2.2⁵⁹

  =2+2.(2²+2⁴+...+2⁵⁹) chia hết cho 2

Vậy A chia hết cho 2

phần b lm tương tự nhé mik ko có thời gian lm tiếp

tích cho mik nha

giúp mình với mình chuẩn bị phải nộp bài rồi T~T 

\(B=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{58}\right)⋮7\)

Lời giải:
$a=1+5+5^2+5^3+...+5^{2022}+5^{2023}$

$5a=5+5^2+5^3+5^4+....+5^{2023}+5^{2024}$

$\Rightarrow 5a-a=5^{2024}-1$

$\Rightarrow 4a=5^{2024}-1$

$\Rightarrow 4a+1=5^{2024}\vdots 5^{2023}$ (đpcm)