Question

Cho N=\(1+2+2^2+2^3+2^4+...+2^{229}+2^{300}\)

\(CMR:N⋮7̸\)

Answer 1

Ta co \(N=1+2+2^2+2^3+2^4+2^5+...+2^{228}+2^{229}+2^{300}\)

\(N=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{228}+2^{229}+2^{300}\right)\)

\(N=\left(1+2+2^2\right)+2^3.\left(1+2+2^2\right)+...+2^{228}.\left(1+2+2^2\right)\)

\(N=\left(1+2+2^2\right).\left(1+2^3+...+2^{228}\right)\)

\(N=7.\left(1+2^3+...+2^{228}\right)\)

Vì \(7⋮7=>7.\left(1+2^3+...+2^{228}\right)⋮7\)

Hay \(N⋮7\)

tick cho mk nha

chưa hiểu chỗ nào thì hỏi

Answer 2

N=\(1+2+2^2+2^3+...+2^{299}+2^{300}\)

N=\(\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{297}+2^{298}+2^{299}\right)+2^{300}\)

N=\(7+2^3\left(1+2+2^2\right)+2^6\left(1+2+2^2\right)+...+2^{297}\left(1+2+2^2\right)+2^{300}\)

N=\(7+2^3.7+2^6.7+...+2^{297}.7+2^{300}\)

N=\(7.\left(1+2^3+2^6+...+2^{297}\right)+2^{300}\)

Ta thấy ​\(7.\left(1+2^3+2^6+...+2^{297}\right)⋮7\)​

Còn lại ta xét số \(2^{300}\).

Khi phân tích ra thừa số nguyên tố thì số \(2^{300}\)không chứa thừa số 7.\(\Rightarrow2^{300}\)không chia hết cho 7

Vậy N không chia hết cho 7.