Question

C = 1+3+3²+...+3¹¹⁹ chia hết cho 13

Answer 1

C = 1 + 3 + 3² + ... + 3¹¹⁹

C = (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

C = 13 + 3³. 13 + ... + 3¹¹⁷.13

C = 13(1 + 3³ + ... + 3¹¹⁷)

Vậy C chia hết cho 13

Answer 2

\(C=1+3+3^2+3^3+3^4+3^5+...+3^{117}+3^{118}+3^{119}\)

\(C=1+3+3^2+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)

\(C=\left(1+3+3^2\right)\left(1+3^3+...+3^{117}\right)\)

\(C=13.\left(1+3^3+...+3^{117}\right)⋮13\)