Question

Cho tổng S= 3+3²+3³+...+3²⁰¹⁹. Chứng minh S chia hết cho 13.

Answer 1

\(S=3+3^2+3^3+...+3^{2019}\)

\(3S=3^2+3^3+...+3^{2019}+3^{2020}\)

\(\Rightarrow3S-S=-3+3^{2020}\)

\(\Rightarrow2S=3^{2020}-3\Rightarrow S=\frac{3^{2020}-3}{2}\)

Ta có: \(S=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2017}+3^{2018}+3^{2019}\right)\)

\(=3\left(1+3+9\right)+3^4\left(1+3+9\right)+...+3^{2017}\left(1+3+9\right)\)

\(=3.13+3^4.13+...+3^{2017}.13\)

\(=13.\left(3+3^4+...+3^{2017}\right)⋮13\)