Question

A=7+7²+7³+...+7²⁰¹⁶

Chứng minh A chia hết cho 57

Answer 1

\(A=7+7^2+7^3+...+7^{2016}\)

\(A=\left(7+7^2+7^3\right)+\left(7^4+7^5+7^6\right)+...+\left(7^{2014}+7^{2015}+7^{2016}\right)\)

\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{2014}\left(1+7+7^2\right)\)

\(A=7.57+7^4.57+...+7^{2014}.57\)

\(A=\left(7+7^4+...+7^{2014}\right).57⋮57\) ( đpcm ) 

Answer 2

Ta có :

\(A=7\left(1+7+7^2\right)+.....+7^{2014}\left(1+7+7^2\right)\)

\(\Rightarrow A=7.57+....+7^{2014}.57\)

\(\Rightarrow A=57.\left(7+....+7^{2014}\right)\)

=> A chia hêt cho 57