Question

Cho A = 1 + 3 + {3^2} + {3^3} + ... + {3^{101}}. Chứng minh rằng A chia hết cho 13  

Answer 1

\(A=1+3+3^2+3^3+...+3^{101}\\=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+...+(3^{99}+3^{100}+3^{101})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+...+3^{99}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+...+3^{99}\cdot13\\=13\cdot(1+3^3+3^6+...+3^{99})\)

Vì \(13\cdot(1+3^3+3^6+...+3^{99})\vdots13\)

nên \(A⋮13\).

Answer 2

A = 1 + 3 + 3² + 3³ + … + 3¹⁰¹

A = (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + … + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)

A = (1 + 3 + 3²) + 3³(1 + 3 + 3²) + … + 3⁹⁹(1 + 3 + 3²)

A = (1 + 3 + 3²)(1 + 3³ + … + 3⁹⁹)

A = 13.(1 + 3³ + … + 3⁹⁹) ⋮ 13

Vậy A chia hết cho 13.