1) \(5+5^2+5^3+.....+5^{12}=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{11}+5^{12}\right)\)

\(=30.1+5^2.30+.....+5^{10}.30=30.\left(1+5^2+....+5^{10}\right)\)

Vậy chia hết cho 30

\(5+5^2+5^3+....+5^{12}=\left(5+5^2+5^3\right)+.....+\left(5^{10}+5^{11}+5^{12}\right)\)

\(=5.31+5^4.31+....+5^{10}.31=31.\left(5+5^4+....+5^{10}\right)\)

Vậy chia hết cho 31

haizzzzzzzzzzz câu 2 làm tek nào z

TA CÓ:

A=3⁰+3+3²+3³+........+3¹¹

(3⁰+3+3²+3³)+....+(3⁸+3⁹+3¹⁰+3¹¹)

3(0+1+3+3²)+......+3⁸(0+1+3+3²) 

3.13+....+3⁸.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)

\(A=17^{18}-17^{16}\\ =17^{16}\cdot\left(17^2-1\right)\\ =17^{16}\cdot\left(289-1\right)\\ =17^{16}\cdot288\\ =17^{16}\cdot18\cdot16⋮18\)

Vậy \(A⋮18\)

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

Ta có: \(52=4\cdot13\)

\(B=1+3+3^2+...+3^{11}\\ =\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\\ =1\cdot\left(1+3\right)+3^2\cdot\left(1+3\right)+...+3^{10}\cdot\left(1+3\right)\\ =\left(1+3\right)\cdot\left(1+3^2+...+3^{10}\right)\\ =4\cdot\left(1+3^2+...+3^{10}\right)⋮4\)

Vậy \(B⋮4\)

\(B=1+3+3^2+...+3^{11}\\ =\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^9+3^{10}+3^{11}\right)\\ =1\cdot\left(1+3+3^2\right)+3^3\cdot\left(1+3+3^2\right)+...+3^9\cdot\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\cdot\left(1+3^3+...+3^9\right)\\ =13\cdot\left(1+3^3+...+3^9\right)⋮13\)

Vậy \(B⋮13\)

Vì \(4\) và \(13\) là hai số nguyên tố cùng nhau nên tao có \(B⋮4\cdot13\Leftrightarrow B⋮52\)

Vậy \(B⋮52\)

\(C=3+3^3+3^5+...3^{31}\)

\(C=3+3^3+3^5+...+3^{31}\\ =\left(3+3^3\right)+\left(3^5+3^7\right)+...+\left(3^{29}+3^{31}\right)\\ =1\cdot\left(3+3^3\right)+3^4\cdot\left(3+3^3\right)+...+3^{28}\cdot\left(3+3^3\right)\\ =\left(3+3^3\right)\cdot\left(1+3^4+...+3^{28}\right)\\ =30\cdot\left(1+3^4+...+3^{28}\right)⋮15\left(\text{vì }30⋮15\right)\)

Vậy \(C⋮15\)

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

Tao có: \(21=3\cdot7;15=3\cdot5\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\\ =2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\\ =\left(1+2\right)\cdot\left(2+2^3+...+2^{59}\right)\\ =3\cdot\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(D⋮3\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{57}+2^{59}\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\\ =2\cdot\left(1+2^2\right)+2^5\cdot\left(1+2^2\right)+...+2^{57}\cdot\left(1+2^2\right)+2^2\cdot\left(1+2^2\right)+...+2^{58}\cdot\left(1+2^2\right)\\ =\left(1+2^2\right)\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)\\ =5\cdot\left(2+2^5+...+2^{57}+2^2+...+2^{59}\right)⋮5\)

Vậy \(D⋮5\)

\(D=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\\ =2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\cdot\left(1+2+2^2\right)\\ =\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{58}\right)\\ =7\cdot\left(2+2^4+...+2^{58}\right)⋮7\)

Ta có:

\(D⋮3;D⋮5\Rightarrow D⋮3\cdot5\Leftrightarrow D⋮15\)

\(D⋮3;D⋮7\Rightarrow D⋮3\cdot7\Leftrightarrow D⋮21\)

Vậy \(D⋮15;D⋮21\)

Mình chỉ làm mẫu 1 câu thui nha:

\(A=17^{18}-17^{16}\)

\(A=17^{16}.17^2-17^{16}.1\)

\(A=17^{16}\left(17^2-1\right)\)

\(A=17^{16}.288\)

\(A=17^{16}.16.18\)

\(A⋮18\left(đpcm\right)\)

B = 1 + 3 + 3² + 3³ + 3⁴ + 3⁵ + ... + 3¹¹

B = (1 + 3 + 3² + 3³ + 3⁴ + 3⁵) + (3⁶ + 3⁷ + 3⁸ + 3⁹ + 3¹⁰ + 3¹¹)

B = 364 + 3⁶.364

B = 364(3⁶ + 1) \(⋮\) 52

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

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

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

\(A=4+3^2\cdot4+....+3^{10}\cdot4\)

\(A=4\cdot\left(1+3^2+...+3^{10}\right)\) chia hết cho 4

Vì ta có 4 chia hết cho 4 => A có chia hết cho 4

Vậy A chia hết cho 4

2:

\(C=5+5^2+5^3+...+5^8\) chia hết cho 30

\(C=\left(5+5^2\right)+...+\left(5^7+5^8\right)\)

\(C=30+5^2\cdot\left(5+5^2\right)+...+5^6\cdot\left(5+5^2\right)\)

\(C=30\cdot1+5^2\cdot30+...5^6\cdot30\)

\(C=30\cdot\left(5^2+...+5^6\right)\)

Vì ta có 30 chia hết cho 30 nên suy ra C có chia hết cho 30

Vậy C có chia hết cho 30

quá dễ

C=(5+5²)+(5³+5⁴)+.......+(5¹¹+5¹²)

=30+5².(5¹+5²)+.....+5¹⁰.(5¹+5²)

=30.1+5².30+.....+5¹⁰.30

=30.(1+5²+.........+5¹⁰) chia hết cho 30

chắc là đúng ahihihi

chịu thoiiiii 

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