\(A=1+3+3^2+..........+3^{11}\)

\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)

\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)

\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)

\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)

A = 1 + 3 + 3² + 3³ + ... + 3¹¹

A = ( 1 + 3 ) + ( 3² + 3³ ) + ... + ( 3¹⁰ + 3¹¹ )

A = 4 + 3² . ( 1 + 3 ) + ... + 3¹⁰ . ( 1 + 3 )

A = 4 + 3² . 4 + ... + 3¹⁰ . 4

A = 4 . ( 1 + 3² + ... + 3¹⁰ ) \(⋮\) 4 ( Vì trong tích có một thừa số chia hết cho 4 )

~ Chúc bạn học giỏi ! ~

1. \(A=2^{2016}-1\)

\(2\equiv-1\left(mod3\right)\\ \Rightarrow2^{2016}\equiv1\left(mod3\right)\\ \Rightarrow2^{2016}-1\equiv0\left(mod3\right)\\ \Rightarrow A⋮3\)

\(2^{2016}=\left(2^4\right)^{504}=16^{504}\)

16 chia 5 dư 1 nên 16^504 chia 5 dư 1

=> 16^504-1 chia hết cho 5

hay A chia hết cho 5

\(2^{2016}-1=\left(2^3\right)^{672}-1=8^{672}-1⋮7\)

lý luận TT trg hợp A chia hết cho 5

(3;5;7)=1 = > A chia hết cho 105

2;3;4 TT ạ !!

=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+.....+(3^2012+3^2013+3^2014+3^2015)

=3(1+3+9+27)+3^5(1+3+9+27)+.....+3^2012(1+3+9+27)

=40(3+3^5+...+3^2012)

=>A chia hết cho 10

a/ \(A=3+3^2+3^3+3^4+.............+3^{49}+3^{50}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+............+\left(3^{49}+3^{50}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+............+3^{49}\left(1+3\right)\)

\(=3.4+3^3.4+...............+3^{49}.4\)

\(=4\left(3+3^3+...........+3^{49}\right)⋮4\)

\(\Leftrightarrow A⋮4\left(đpcm\right)\)

b/ \(A=3+3^2+3^3+3^4+.............+3^{49}+3^{50}\)

\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^9\right)+........+\left(+3^{47}+3^{48}+3^{49}+3^{50}\right)\)

\(=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+........+3^{47}\left(1+3+3^2+3^3\right)\)

\(=3.40+3^5.40+.........+3^{47}.40\)

\(=40\left(3+3^5+...........+3^{47}\right)⋮10\)

\(\Leftrightarrow A⋮10\left(đpcm\right)\)

Bạn lấy 1 và 3, 2 và 4, 5 và 7....48 và 50 cộng với nhau có tổng chia hết cho 10 Suy ra a chia hết cho 10

Bài 1 :

\(3^{22}-9^{10}-27^6=3^{22}-\left(3^2\right)^{10}-\left(3^3\right)^6=3^{22}-3^{20}-3^{18}=3^{18}.\left(3^4-3^2-1\right)=3^{18}.71\)chia hết cho 71 (đpcm).