Question

[RPK Svip]

\(3^{n+3}\) + \(3^{n+1}\) + \(2^{n+2}\) + \(2^{n+1}\)

Chứng minhA chia hết cho 6

Answer 1

[Thảo Nguyễn『緑』]

\(3^{n+3}+3^{n+1}+2^{n+2}+2^{n+1}\)|

\(=3^n\cdot3^3+3^n\cdot3+2^n\cdot2^2+2^n\cdot2\)

\(=3^n\left(3^3+3\right)+2^n\left(2^2+2\right)\)

\(=3^n\cdot30+2^n\cdot6\)

Vì 30 chia hết cho 6 nên 3ⁿ . 30 cũng chia hết cho 6.

Vì 6 chia hết cho 6 nên 2ⁿ .6 cũng chia hết cho 6.

Vậy .....

=))

Answer 2

[응 우옌 민 후엔]

Ta có: 

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

   \(=3^{n+1}\cdot3^2+3^{n+1}+2^{n+1}\cdot2^1+2^{n+1}\)

   \(=3^{n+1}\cdot\left(3^2+1\right)+2^{n+1}\cdot\left(2^1+1\right)\)

   \(=3^{n+1}\cdot10+2^{n+1}\cdot3\)

   \(=3^n\cdot3\cdot2\cdot5+2^n\cdot2\cdot3\)

   \(=3^n\cdot6\cdot5+2^n\cdot6\)

   \(=6\cdot\left(3^n\cdot5\cdot2^n\right)\Rightarrow⋮6\left(đpcm\right)\)