Question

[Phạm Kiều Trang]

chứng tỏ a+3^1+3^2+3^3+...+3^60 chia hết chi 13

Answer 1

[HT.Phong (9A5)]

\(A=3^1+3^2+...+3^{60}\)

\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(A=3\cdot\left(1+3+3^2\right)+3^4\cdot\left(1+3+3^2\right)+...+3^{58}\cdot\left(1+3+3^2\right)\)

\(A=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{58}\cdot\left(1+3+9\right)\)

\(A=3\cdot13+3^4\cdot13+...+3^{58}\cdot13\)

\(A=13\cdot\left(3+3^4+...+3^{58}\right)\)

Mà: \(13\cdot\left(3+3^4+...+3^{58}\right)\) ⋮ 13

\(\Rightarrow A\) ⋮ 13