chứng minh theo phương pháp quy nạp: 6^2n+3^(n+2)+3^n chia hết cho 11

$$\eqalign{ & {6^{2n}} + {3^{n + 2}} + {3^n} \cr & Voi\,\,n = 0:\,\,{6^0} + {3^2} + {3^0} = 11\,\, \vdots \,\,11 \cr & Gia\,\,su\,\,{6^{2k}} + {3^{k + 2}} + {3^k}\,\, \vdots \,\,11 \cr & Ta\,\,chung\,\,\min h\,\,{6^{2\left( {k + 1} \right)}} + {3^{k + 1 + 2}} + {3^{k + 1}}\,\, \vdots \,\,11 \cr & Ta\,\,co:\,\, \cr & \,\,\,\,{6^{2\left( {k + 1} \right)}} + {3^{k + 1 + 2}} + {3^{k + 1}} \cr & = {36.6^{2k}} + {3.3^{k + 2}} + {3.3^k} \cr & = {33.6^{2k}} + 3\left( {{3^{2k}} + {3^{k + 2}} + {3^k}} \right) \cr & \left\{ \matrix{ 33\,\, \vdots \,\,11 \Rightarrow {33.6^{2k}}\,\, \vdots \,\,11 \hfill \cr {3^{2k}} + {3^{k + 2}} + {3^k}\,\, \vdots \,\,11 \Rightarrow 3\left( {{3^{2k}} + {3^{k + 2}} + {3^k}} \right)\,\, \vdots \,\,11 \hfill \cr} \right. \cr & \Rightarrow {3.6^{2k}} + 3\left( {{3^{2k}} + {3^{k + 2}} + {3^k}} \right) \vdots \,\,11 \cr & \Rightarrow \,{6^{2\left( {k + 1} \right)}} + {3^{k + 1 + 2}} + {3^{k + 1}}\,\, \vdots \,\,11 \cr & Vay\,\,{6^{2n}} + {3^{n + 2}} + {3^n}\,\, \vdots \,\,11\,\,\forall n \ge 0 \cr} $$