Question

Cho S = 1 + 2 + 2² + 2³ + ..... + 2²⁰¹⁸. Tìm số dư khi S chia cho 7

Answer 1

Ta có : \(S=1+2+2^2+2^3+...+2^{2018}\)

= \(\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...\left(2^{2016}+2^{2017}+2^{2018}\right)\)

= \(\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...2^{2016}\left(1+2+2^2\right)\)

= \(\left(1+2+2^2\right)\left(1+2^3+2^6+...2^{2016}\right)\)

= \(7\left(1+2^3+2^6+...+2^{2016}\right)\)\(⋮7\)

Vậy S:7 dư 0