ko bt lm

a) 4.(1+4)+4³.(1+4)+................+4⁵⁹(1+4)

=5.4+5.4³+...+5.4⁵⁹

=5.(4+4³+.+4⁵⁹) chia hết cho 5

4.(1+4+4²)+4⁴.(1+4+4²)+...............+4⁵⁸(1+4+4²)

=21.4+4⁴.21+..+21.4⁵⁸

=21.(4+4⁴+.+4⁵⁸) chia hết cho 21

b) 5.(1+5)+5³(1+5)+.+5⁹(1+5)

=6.(5+5³+.............+5⁹) chia hết cho 6

a) Đặt biểu thức trên là A, ta có:

A = 4 + 4² + 4³ + 4⁴ + ... + 4⁶⁰

=> A = (4 + 4²) + (4³ + 4⁴) + ... + (4⁵⁹ + 4⁶⁰)

=> A = 4(1 + 4) + 4³(1 + 4) + ... + 4⁵⁹(1 + 4)

=> A = 4 . 5 + 4³ . 5 + ... + 4⁵⁹ . 5

=> A = 5(4 + 4³ + ... + 4⁵⁹)

=> A ⋮ 5

A = 4 + 4² + 4³ + 4⁴ + ... + 4⁶⁰

=> A = (4 + 4² + 4³) + (4⁴ + 4⁵ + 4⁶) + ... + (4⁵⁸ + 4⁵⁹ + 4⁶⁰)

=> A = 4(1 + 4 + 4²) + 4⁴(1 + 4 + 4²) + ... + 4⁵⁸(1 + 4 + 4²)

=> A = 4 . 21 + 4⁴ . 21 + ... + 4⁵⁸ . 21

=> A = 21(4 + 4⁴ + ... + 4⁵⁸)

=> A ⋮ 21

b) Đặt biểu thức trên là B, ta có:

B = 5 + 5² + 5³ + 5⁴ + ... + 5¹⁰

=> B = (5 + 5²) + (5³ + 5⁴) + ... + (5⁹ + 5¹⁰)

=> B = 5(1 + 5) + 5³(1 + 5) + ... + 5⁹(1 + 5)

=> B = 5 . 6 + 5³ . 6 + ... + 5⁹ . 6

=> B = 6(5 + 5³ + ... + 5⁹)

=> B ⋮ 6  

Bài 3:

\(A=5+5^2+..+5^{12}\)

\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)

\(5A=5^2+5^3+...+5^{13}\)

\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)

\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)

\(4A=5^{13}-5\)

\(A=\dfrac{5^{13}-5}{4}\)

a, 4 + \(4^2\) + \(4^3\) + ... + \(4^{60}\) chia hết cho 5

= ( 4 + \(4^2\) ) + ( \(4^3\) + \(4^4\) ) +... + ( \(4^{59}\) + \(4^{60}\))

= ( 4 + \(4^2\) ) + \(4^3\) . ( 4 + \(4^2\) ) +... + \(4^{59}\). ( 4 + \(4^2\) )

= 20 + \(4^3\) . 20 + ... + \(4^{59}\) . 20

= 20 . ( 1 + \(4^3\) + ... + \(4^{59}\) ) chia hết cho 5

4 + \(4^2\) + \(4^3\) + ... + \(4^{60}\) chia hết cho 21

= ( 4 + \(4^2\) + \(4^3\) ) + ( \(4^4\) + \(4^5\) + \(4^6\) ) + ... + ( \(4^{58}\)+ \(4^{59}\) + \(4^{60}\) )

= ( 4 + \(4^2\) + \(4^3\) ) + \(4^4\) . ( 4 + \(4^2\) + \(4^3\) ) + ... + \(4^{58}\) . ( 4 + \(4^2\) + \(4^3\) )

= 84 + \(4^4\) . 84 + .... + \(4^{58}\) . 84

= 84 . ( 1 + \(4^4\) + ... + \(4^{58}\) ) chia hết cho 21

b, 5 + \(5^2\) + \(5^3\) + ... + \(5^{10}\) chia hết cho 6

= ( 5 + \(5^2\) ) + ( \(5^3\) + \(5^4\) ) + ... + ( \(5^9\) + \(5^{10}\) )

= ( 5 + \(5^2\) ) + \(5^3\) . ( 5 + \(5^2\) ) + ... + \(5^9\) . ( 5 + \(5^2\) )

= 30 + \(5^3\) . 30 + ... + \(5^9\) . 30

= 30 . ( 1 + \(5^3\) + ... + \(5^9\) ) chia hết cho 6

c)D=4+4²+4³+4⁴+...+4²⁰¹²

D=(4+4²)+(4³+4⁴)+...+(4²⁰¹¹+4²⁰¹²)

D=4.5+4³.5+4⁵.5+...+4²⁰¹¹.5

D=5.(4+4³+4²⁰¹¹)

=>D chia hết cho 5

=>ĐPCM

tất cả đều có trong câu hỏi tương tự

\(A=1+4+4^2+...+4^{2012}=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{2010}\left(1+4+4^2\right)\)

\(=21+21.4^3+...+21.4^{2010}=21\left(1+4^3+...+4^{2010}\right)⋮21\)

\(B=1+7+7^2+...+7^{101}=\left(1+7\right)+7^2\left(1+7\right)+...+7^{100}\left(1+7\right)\)

\(=8+7^2.8+...+7^{100}.8=8\left(1+7^2+...+7^{100}\right)⋮8\)

Bài 5: 

b: Ta có: \(n+6⋮n+2\)

\(\Leftrightarrow n+2\in\left\{2;4\right\}\)

hay \(n\in\left\{0;2\right\}\)

c: Ta có: \(3n+1⋮n-2\)

\(\Leftrightarrow n-2\in\left\{-1;1;7\right\}\)

hay \(n\in\left\{1;3;9\right\}\)