Đáp án:
$\begin{array}{l}
a)2n - 3\\
= 2n + 2 - 5\\
= 2.\left( {n + 1} \right) - 5\\
Do:2.\left( {n + 1} \right) \vdots \left( {n + 1} \right)\\
\Rightarrow 5 \vdots \left( {n + 1} \right)\\
\Rightarrow \left[ \begin{array}{l}
n + 1 = 1\\
n + 1 = 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
n = 1 - 1 = 0\\
n = 5 - 1 = 4
\end{array} \right.\\
Vậy\,n = 0;n = 4\\
b)Do:3n \vdots n\\
\Rightarrow 5 \vdots n\\
\Rightarrow \left[ \begin{array}{l}
n = 1\\
n = 5
\end{array} \right.\\
Vậy\,n = 1;n = 5\\
c)A = 1 + 5 + {5^2} + ... + {5^{204}} + {5^{205}} + {5^{206}}\\
= \left( {1 + 5 + {5^2}} \right) + \left( {{5^3} + {5^4} + {5^5}} \right) + ... + \left( {{5^{204}} + {5^{205}} + {5^{206}}} \right)\\
= 31 + {5^3}.\left( {1 + 5 + {5^2}} \right) + ... + {5^{204}}.\left( {1 + 5 + {5^2}} \right)\\
= 31 + {5^3}.31 + ... + {5^{204}}.31\\
= \left( {1 + {5^3} + ... + {5^{204}}} \right).31 \vdots 31\\
Vậy\,A \vdots 31
\end{array}$
