Question

Cho A = \(^{4^n+4^{n+1}+4^{n+2}+4^{n+3}+...+4^{n+20}}\)( với n\(\varepsilon\)N*)

Chứng minh rằng A chia hết cho 84

Answer 1

\(A=4^{n-1}\left(4+4^2+4^3\right)+4^{n+3}\left(4+4^2+4^3\right)+...+4^{n+17}\left(4+4^2+4^3\right)\)

\(\Rightarrow A=4^{n-1}\times84+4^{n+3}\times84+...+4^{n+17}\times84\)

\(\Rightarrow A=84\left(4^{n-1}+4^{n+3}+...+4^{n+17}\right)⋮84\)

Vậy \(A⋮84\) 

Answer 2

Yêu cầu bài này là gì vậy bạn ơi ?

Answer 3

Mk đăng r đó !

Answer 4

\(A=\left(4^n+4^{n+1}+4^{n+2}\right)+\left(4^{n+3}+4^{n+4}+4^{n+5}\right)+...+\left(4^{n+18}+4^{n+19}+4^{n+20}\right).\)

\(=4^{n-1}\left(4+4^2+4^3\right)+4^{n+1}\left(4+4^2+4^3\right)+..+4^{n+19}\left(4+4^2+4^3\right)\)

\(=84\left(4^{n-1}+4^{n+1}+...+4^{n+19}\right)⋮84\)

Vậy A chia hết cho 84