Question

ai giúp mình câu này với

Cho A=3+3^2+3^3+...+3^23+3^24. Chứng minh rằng A chia hết cho 12

Giúp mình nha❤

Answer 1

\(A=3+3^2+...+3^{24}\\ =\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{23}+3^{24}\right)\\ =12+3^2\left(3+3^2\right)+...+3^{22}\left(3+3^2\right)\\ =12+3^2.12+...+12.3^{22}\\ =12\left(1+...+3^{22}\right)⋮12\)

Answer 2

A = 3 + 3² + 3³ + ......... + 3²³ + 3²⁴

A = ( 3 + 3² ) + ( 3³ + 3⁴ ) + ............ + ( 3²³ + 3²⁴ )

A = 12 + 3² . ( 3 + 3² ) + .............. + 3²³ . ( 3 + 3² )

A = 12 + 3² . 12 + ............... + 3²³ . 12

A = 12 . ( 1 + 3² + ............. + 3²³ )

Mà 12 \(⋮\) 12 \(\Rightarrow\) 12 . ( 1 + 3² + .............. + 3²³ ) \(⋮\) 12

Vậy A \(⋮\) 12 ( ĐPCM )

Answer 3

Ta có:3+3^2+3^3+...+3^24(gồm 24 số hạng)

<=>(3+3^2)+(3^3+3^4)+...+(3^23+3^24)

<=>1(3+3^2)+3^2(3+3^2)+...+3^22(3+3^2)

<=>1.12+3^2.12+...+3^22.12

<=>12(1+3^2+3^4+...+3^24)

Vậy A chia hết cho 12