Question

chứng minh rằng:

E=1+3+3^2+3^3+...+3^119  chia hết cho 13

giúp tui nha

Answer 1

E = 1 + 3 + 3² + 3³ +.....+3¹¹⁹

E = ( 1 + 3 + 3²) +....+ ( 3¹¹⁷ + 3¹¹⁸+ 3¹¹⁹)

E =   13 + ......+ 3¹¹⁷.( 1 + 3 + 3²)

E = 13 +.....+ 3¹¹⁷ . 13

E = 13. ( 3⁰ + ....+ 3¹¹⁷)

13 ⋮ 13 ⇒ 13. (3⁰ +....+3¹¹⁷) ⋮ 13 ⇒ E = 1 +3+3²+ ....+3¹¹⁹⋮13(đpcm)

Answer 2

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

= \(13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)

=\(13+3^3.13+...+3^{117}.13\)

=\(13.\left(1+3^2+...+3^{117}\right)\) chia hết cho 13