Question

cho M= 4 + 4^2 + 4^3 + 4^4 + ... +4^2022. chứng minh rằng M chia hết cho 21. mọi người giúp mình với ạ

Answer 1

\(M=4+4^2+4^3+4^4+...+4^{2022}\)

\(=\left(4+4^2+4^3\right)+\left(4^4+4^5+4^6\right)+...+\left(4^{2020}+4^{2021}+4^{2022}\right)\)

\(=4.\left(1+4+4^2\right)+4^4.\left(1+4+4^2\right)+...+4^{2020}.\left(1+4+4^2\right)\)

\(=4.21+4^4.21+...+4^{2020}.21\)

\(=21.\left(4+4^4+...+4^{2020}\right)\)

Vì \(21⋮21=>M⋮21\)

Answer 2

M=4(1+4+4^2)+4^4(1+4+4^2)+...+4^2020(1+4+4^2)

=21(4+4^4+...+4^2020) chia hết cho 21