Giải thích các bước giải:
Ta có:
$A=3^{n+3}-2^{n+3}+3^{n+1}-2^{n+1}$
$\to A=(3^{n+3}+3^{n+1})-(2^{n+3}+2^{n+1})$
$\to A=3^{n+1}(3^{2}+1)-2^{n+1}(2^{2}+1)$
$\to A=3^{n+1}\cdot 10-2^{n+1}\cdot 5$
Ta có: $2^{n+1}\cdot 5\quad\not\vdots\quad 3$
$\to 3^{n+1}\cdot 10-2^{n+1}\cdot 5\quad\not\vdots\quad 3$
$\to 3^{n+3}-2^{n+3}+3^{n+1}-2^{n+1}\quad\not\vdots\quad 3$
$\to 3^{n+3}-2^{n+3}+3^{n+1}-2^{n+1}\quad\not\vdots\quad 6$