Question

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

a) Chứng minh A chia hết cho 4

b) Chứng minh A chia hết cho 5

c) Chứng minh A chia hết cho 21

Answer 1

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

b) \(A=4+4^2+4^3+...+4^{60}=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{59}\left(1+4\right)=4.5+4^3.5+...+4^{59}.5=5\left(4+4^3+...+4^{59}\right)⋮5\)

c) \(A=4+4^2+4^3+...+4^{60}=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+...+4^{58}\left(1+4+4^2\right)=4.21+4^4.21+...+4^{58}.21=21\left(4+4^4+...+4^{58}\right)⋮21\)

Answer 2

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

\(A=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+....4^{57}.\left(1+4+4^2\right)\)

\(A\)\(=21+4^3.21+...4^{57}.21\)

\(\Rightarrow A⋮4;21\)

ko chia hết cho 5

 

Answer 3

a:Ta có: \(A=4+4^2+4^3+...+4^{60}\)

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

b: Ta có: \(A=4+4^2+4^3+...+4^{60}\)

\(=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{59}\left(1+4\right)\)

\(=5\cdot\left(4+4^3+...+4^{59}\right)⋮5\)

Answer 4

Trả lời:

= 1 + (-5) + (-5)² + (-5)³ + ... + (-5)¹⁰⁰

A - (-5).A = \(\left[1+\left(-5\right)+\left(-5\right)^2+....+\left(-5\right)^{100}\right]-\left[-5+\left(-5\right)^2+...+\left(-5\right)^{101}\right]\)

=> A.(-5 - 1) = (-5)¹⁰¹ - 1

=> \(\left[1-\left(-5\right)\right]\) = 1 - (-5)¹⁰¹

6A = 1 + 5¹⁰¹

=> A = \(\dfrac{1+5^{101}}{6}\)