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)

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

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

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

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

M=1+3+3^2+3^3+^3+...+3^118+3^119

  =(1+3+3^2)+(3^3+3^4+3^5)+...+(3^117+3^118+3^119)

 =13+3^3(1+3+3^2)+...+3^117(1+3+3^2)

 =13+3^3.13+..+3^117.13

 =13(1+3^3+...+3^117) chia hết cho 13

Vậy Mchia hết cho 13

ai chơi truy kích thì kết bạn vs mình nha 

rồi khi nào tạo phòng solo đao phong chibi với nhau 1 ván

ai chơi truy kích 

kb với mình mình k cho mình chưa có bạn

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

=> A=13+3³(1+3+3²)+...+3²⁰¹⁷(1+3+3²) 

=> A=13+3³.13+...+3²⁰¹⁷.13 

=> A=13(1+3³+...+3²⁰¹⁷) chia hết cho 13

\(B=3+3^2+3^3+...+3^{118}+3^{119}+3^{120}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\\ =3.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\\ =\left(3+3^4+...+3^{118}\right).\left(1+3+3^2\right)\\ =\left(3+3^4+...+3^{118}\right).13⋮13\left(ĐPCM\right)\)

thanks

Hứa r phải làm thôi:

Đặt:

\(A=1+3+3^2+.....+3^{119}\)

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

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

\(A=1.13+3^3.13+....+3^{117}.13\)

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

\(A⋮13\rightarrowđpcm\)

Đặt \(A=1+3+3^2+3^3+...+3^{119}\)

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

\(=13+\left(1.3^3+3.3^3+3^2.3^3\right)+...+\left(1.3^{117}+3.3^{117}+3^2.3^{117}\right)\)

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

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

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

Vậy \(A⋮13\)

giúp mình với mình chuẩn bị phải nộp bài rồi T~T 

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

\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{58}\right)⋮7\)

\(D=1+3+3^2+...+3^{119}\)

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

\(=13\left(1+...+3^{117}\right)⋮13\)

A = ( 1 + 3 + 3² ) + 3³( 1 + 3 + 3² ) + ... + 3¹¹⁷( 1 + 3 + 3² )

A = 13 + 3³ . 13 + 3⁶ . 13 + ... + 3¹¹⁷ . 13

A = 13 ( 1 + 3³ + 3⁶ + ... + 3¹¹⁷ ) chia hết cho 13 ( vì 13 chia hết cho 13 )

Vậy A chia hết cho 13