Question

cho S=\(3^0+3^1+3^2+...+3^{2018}\)\(CMR\left(S-4\right)⋮39\)

GIÚP MÌNH GẤP MAI NỘP RỒI

 

Answer 1

Ta có : S = 1 + 3¹ + 3² + .... + 3²⁰¹⁸

=> S - 4 =  1 + 3¹ + 3² + .... + 3²⁰¹⁸ - 4

=> S - 4 = 3² + 3³ + 3⁴ + ..... + 3²⁰¹⁸

=> S - 4 = (3² + 3³ + 3⁴ ) + ...... + (3²⁰¹⁶ + 3²⁰¹⁷ + 3²⁰¹⁸)

=> S - 4 = 3(3 + 3² + 3³) + ..... + 3²⁰¹⁵(3 + 3² + 3³)

=> S - 4 = 3.39 + .... + 3²⁰¹⁵.39

=> S - 4 = 39 (3 + .... + 3²⁰¹⁵) chia hết cho 39

Answer 2

Ta thấy S=(3S-S):2

S=3^0+3^1+3^2+...+3^2018

\(\Rightarrow\)3S=3+3^2+3^3+...+3^2019

\(\Rightarrow\)3S-S=(3+3^2+3^3+..+3^2019)-(3^0+3^1+3^2+...+3^2018)

\(\Rightarrow\)3S-S=3^2019-3^0=3^2019-1\(\Rightarrow\)conf thiếu để bên dưới