$\begin{align}  & \text{To show that 17 is a factor of }{{3}^{4n+2}}+2\cdot {{4}^{3n+1}} \\ & \text{Let }n=1 \\ & {{3}^{4(1)+2}}+2\cdot {{4}^{3(1)+1}}={{3}^{6}}+2\cdot {{4}^{4}}=729+512=1241=17(73) \\ & \text{Let }n=2 \\ & {{3}^{4(2)+2}}+2\cdot {{4}^{3(2)+1}}={{3}^{10}}+2\cdot {{4}^{7}}=59049+32768=91817=17(5401) \\ & \text{The proposition holds true for }n=\text{ }1,2 \\ & \text{Assuming the preposition is true for }n=k,\text{ the preposition becomes }{{3}^{4k+2}}+2\cdot {{4}^{3k+1}} \\ & \text{Meaning that }{{3}^{4k+2}}+2\cdot {{4}^{3k+1}}=17M \\ & {{3}^{4k+2}}+2\cdot {{4}^{3k+1}}=17M \\ & 2\cdot {{4}^{3k+1}}=17M-{{3}^{4k+2}} \\ & {{4}^{3k+1}}=\frac{17M-{{3}^{4k+2}}}{2}------(a) \\ & \text{Hence we need to show that the preposition is valid for }k+1 \\ & {{3}^{4(k+1)+2}}+2\cdot {{4}^{3(k+1)+1}} \\ & {{3}^{4k+4+2}}+2\cdot {{4}^{3k+3+1}} \\ & {{3}^{4k+6}}+2\cdot {{4}^{3k+1+3}} \\ & {{3}^{4k+6}}+2\cdot ({{4}^{3k+1}})\cdot {{4}^{3}} \\ & {{3}^{4k+6}}+2\cdot ({{4}^{3k+1}})\cdot 64 \\ & {{3}^{4k+6}}+128({{4}^{3k+1}}) \\ & \text{substituting }{{4}^{3k+1}}=\frac{17M-{{3}^{4k+2}}}{2}\text{ from equation }(a) \\ & {{3}^{4k+6}}+128\left( \frac{17M-{{3}^{4k+2}}}{2} \right) \\ & {{3}^{4k+6}}+64(17M-{{3}^{4k+2}}) \\ & {{3}^{4k}}\cdot {{3}^{6}}+1088-64\cdot ({{3}^{4k+2}}) \\ & ({{3}^{4k}}\times {{3}^{6}})+1088-64\cdot ({{3}^{4k}}\times {{3}^{2}}) \\ & 729\times {{3}^{4k}}+1088-576({{3}^{4k}}) \\ & 729({{3}^{4k}})-576({{3}^{4k}})+1088 \\ & {{3}^{4k}}(729-576)+1088 \\ & {{3}^{4k}}(153)+1088 \\ & 17[{{3}^{4k}}(9)+64] \\ & 17[{{3}^{4k+2}}+64] \\ & Hence,\text{ }this\text{ }shows\text{ }that\text{ }17\text{ }is\text{ }a\text{ }factor\text{ }of\text{ }{{3}^{4n+2}}+2\cdot {{4}^{3n+1}} \\\end{align}$