= (1 + 3 + 3^2) + ....... + (3^2013 + 3^2014+  3^2015)

=1.13 + ...... + 3^2013.13

=13(1 + 3^3 + ... + 3^2013) 

=> chia hết cho 13

\(\text{Đặt }A=1+3+3^2+...+3^{2015}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{2013}+3^{2014}+3^{2015}\right)\)

\(=\left(1+3+9\right)+3^3.\left(1+3+9\right)+...+3^{2013}.\left(1+3+9\right)\)

\(=13+3^3.13+...+3^{2013}.13\)

\(=13.\left(1+3^3+...+3^{2013}\right)\text{chia hết cho 13}\)

=> A chia hết cho 13 (đpcm).

Bạn nhóm 3 số lại 

A=1+3+3²+3³+....+3²⁰¹⁴+3²⁰¹⁵

A=1+(3+3²+3³)+......+(3²⁰¹³+3²⁰¹⁴+3²⁰¹⁵)

A=1+3(1+3+3²)+......+3²⁰¹³+(1+3+3²)

A=1+(3.13)+.....+(3²⁰¹³+13)

A=13.(1+3+....+3²⁰¹³⁾

^{SUY RA : A CHIA HET CHO 13}

\(A=3+3^2+...+3^{2016}\)

\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2015}+3^{2016}\right)\)

\(A=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{2015}\cdot\left(1+3\right)\)

\(A=4\cdot\left(3+3^3+...+3^{2015}\right)\)

Vậy A chia hết cho 4

_____________

\(A=3+3^2+3^3+...+3^{2016}\)

\(A=\left(3+3^2+3^3\right)+...+\left(3^{2014}+3^{2015}+3^{2016}\right)\)

\(A=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{2014}\cdot\left(1+3+9\right)\)

\(A=13\cdot\left(3+3^4+...+3^{2014}\right)\)

Vậy A chia hết cho 13

a) A = 2014 + 2014² + 2014³ + 2014⁴ + ..... + 2014²⁰¹⁴

A = ( 2014 + 2014² ) + ( 2014³ + 2014⁴ ) + ..... + ( 2014²⁰¹³ + 2014²⁰¹⁴ )

A = 2014( 1 + 2014 ) + 2014³( 1 + 2014 ) + ....... 2014²⁰¹³( 1 + 2014 )

A = 2014 . 2015 + 2014³ . 2015 + ....... + 2014²⁰¹³ . 2015

A = ( 2014 + 2014³ + ...... 2014²⁰¹³ ) . 2015 chia hết cho 2015

b) Ta có 6 chia hết cho n - 1

=> n-1 thuộc Ư(6) = { 1 ; 2 ; 3 ; 6 }

Nếu n - 1 = 1 => n = 2 (tm)

Nếu n - 1 = 2 => n = 3 (tm)

Nếu n - 1 = 3 => n = 4 (tm)

Nếu n - 1 = 6 => n = 7 (tm)

Vậy n thuộc { 2 ; 3 ; 4 ; 7 }

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