Đáp án:
Giải thích các bước giải:
${{11}^{n+2}}+{{12}^{2n+1}}\,\,\,\vdots \,\,\,133$
$\bullet \,\,\,\,\,{{11}^{n+2}}+{{12}^{2n+1}}$
$={{11}^{n}}{{.11}^{2}}+{{12}^{2n}}.12$
$={{11}^{n}}.121+{{\left( {{12}^{2}} \right)}^{n}}.12$
$={{11}^{n}}.121+{{144}^{n}}.12$
$={{11}^{n}}\left( 133-12 \right)+{{144}^{n}}.12$
$={{11}^{n}}.133-{{11}^{n}}.12+{{144}^{n}}.12$
$={{11}^{n}}.133+12\left( {{144}^{n}}-{{11}^{n}} \right)$
$\bullet \,\,\,$Ta có công thức ${{a}^{n}}-{{b}^{n}}\,\,\,\vdots \,\,\,\left( a-b \right)$, áp dụng công thức đó để giải bài toán này:
$\bullet \,\,\,$Ta thấy:
${{144}^{n}}-{{11}^{n}}\,\,\,\vdots \,\,\,\left( 144-11 \right)$
$\to {{144}^{n}}-{{11}^{n}}\,\,\,\vdots \,\,\,133$
$\to 12\left( {{144}^{n}}-{{11}^{n}} \right)\,\,\,\vdots \,\,\,133$
Mà ${{11}^{n}}.133\,\,\,\vdots \,\,\,133$
$\to {{11}^{n}}.133+12\left( {{144}^{n}}-{{11}^{n}} \right)\,\,\,\vdots \,\,\,133$
$\to {{11}^{n+2}}+{{12}^{2n+1}}\,\,\,\vdots \,\,\,133$
