Cho `n` là số tự nhiên Chứng minh: `5^(2n + 1) + 2^(n + 4) + 2^(n:1)`chia hết cho `23`

Đáp án:  ${{5}^{2n+1}}+{{2}^{n+4}}+{{2}^{n+1}}\,\,\,\vdots 23$

Giải thích các bước giải:

${{5}^{2n+1}}+{{2}^{n+4}}+{{2}^{n+1}}$

$={{5.5}^{2n}}+{{2}^{4}}{{.2}^{n}}+{{2.2}^{n}}$

$={{5.25}^{n}}+{{18.2}^{n}}$

$={{5.25}^{n}}+{{23.2}^{n}}-{{5.2}^{n}}$

$={{23.2}^{n}}+5\left( {{25}^{n}}-{{2}^{n}} \right)$

Ta có ${{25}^{n}}-{{2}^{n}}\,\,\,\vdots\,\,\,\left(25-2\right)$

$\to {{25}^{n}}-{{2}^{n}}\,\,\,\vdots\,\,\,23$

$\to 5\left( {{25}^{n}}-{{2}^{n}} \right)\,\,\,\vdots\,\,\,23$

Ta có $23\,\,\,\vdots\,\,\,23$

$\to {{23.2}^{n}}\,\,\,\vdots\,\,\,23$

Vậy ${{23.2}^{n}}+5\left( {{25}^{n}}-{{2}^{n}} \right)\,\,\,\vdots\,\,\,23$

$\to {{5}^{2n+1}}+{{2}^{n+4}}+{{2}^{n+1}}\,\,\,\vdots\,\,\,23$