Question

[Herera Scobion]

chứng minh 3²ⁿ - 9 chia hết cho 72

Answer 1

[👁💧👄💧👁]

Có: \(3^{2n}-9=\left(3^n\right)^2-3^2=\left(3^n-3\right)\left(3^n+3\right)\)

Có: \(\left\{{}\begin{matrix}3^n-3⋮3\\3^n+3⋮3\end{matrix}\right.\)

\(\Rightarrow\left(3^n-3\right)\left(3^n+3\right)⋮9\)

Lại có: \(\left\{{}\begin{matrix}3^n-3⋮2\\3^n+3⋮2\end{matrix}\right.\)( vì cả 2 số đều là số chẵn)

+ Nếu \(3^n+3\) chia 4 dư 2 thì \(3^n-3⋮4\)

\(\Rightarrow\left(3^n-3\right)\left(3^n+3\right)⋮4\cdot2=8\)

+ CMTT trên, nếu \(3^n+3⋮4\) thì \(\left(3^n-3\right)\left(3^n+3\right)⋮8\)

Vậy \(\left(3^n-3\right)\left(3^n+3\right)⋮8\)

Mà \(\left(8;9\right)=1\)

\(\Rightarrow\left(3^n-3\right)\left(3^n+3\right)⋮8\cdot9=72\\ \Leftrightarrow3^{2n}-9⋮72\left(đpcm\right)\)