$\begin{align}  & \text{To show that 7 is a factor of }{{4}^{3n-1}}+{{2}^{3n-1}}+1 \\ & \text{When }n=1 \\ & {{4}^{3(1)-1}}+{{2}^{3(1)-1}}+1={{4}^{2}}+{{2}^{2}}+1=21=7(3) \\ & \text{When }n=2 \\ & {{4}^{3(2)-1}}+{{2}^{3(2)-1}}+1={{4}^{5}}+{{2}^{5}}+1=21=7(151) \\ & \text{Hence, 7 is a factor of }{{4}^{3n-1}}+{{2}^{3n-1}}+1\text{ since it holds true }n=1,2 \\ & \text{Assuming that the preposition hold true for }n=k \\ & \text{Then, the preposition hold true for }{{4}^{3k-1}}+{{2}^{3k-1}}+1 \\ & {{4}^{3k-1}}+{{2}^{3k-1}}+1=7(M) \\ & {{4}^{3k-1}}=7(M)-{{2}^{3k-1}}-1------(a) \\ & \text{We then wish to show that it holds true for }n=k+1 \\ & \text{for }n=k+1 \\ & {{4}^{3(k+1)-1}}+{{2}^{3(k+1)-1}}+1 \\ & {{4}^{3k+3-1}}+{{2}^{3k+3-1}}+1 \\ & {{4}^{3k-1+3}}+{{2}^{3k+2}}+1 \\ & {{4}^{3k-1}}\cdot {{4}^{3}}+{{2}^{3k+2}}+1 \\ & \text{Substituting }7(M)-{{2}^{3k-1}}-1\text{ for }{{4}^{3k-1}}\text{ } \\ & \text{in the above expression} \\ & [7(M)-{{2}^{3k-1}}-1\text{  }\!\!]\!\!\text{ }{{\text{4}}^{3}}+{{2}^{3k+2}}+1 \\ & 448M-64({{2}^{3k-1}})-64+{{2}^{3k+2}}+1 \\ & 448M-63-{{2}^{6}}\cdot ({{2}^{3k-1}})+{{2}^{3k+2}} \\ & 448M-63-{{2}^{3k-1+6}}+{{2}^{3k+2}} \\ & 448M-63-{{2}^{3k+5}}+{{2}^{3k+2}} \\ & 448M-63-{{2}^{3k}}\cdot {{2}^{5}}+{{2}^{3k}}\cdot {{2}^{2}} \\ & 448M-63-{{2}^{3k}}({{2}^{5}}-{{2}^{2}}) \\ & 448M-63-{{2}^{3k}}(28) \\ & 7(64M-9-{{2}^{3k}}\cdot 4) \\ & 7(64M-9-{{2}^{3k}}\cdot {{2}^{2}}) \\ & 7(64M-9-{{2}^{3k+2}}) \\ & Hence,this\text{ show that }7\text{ is a factor of }{{4}^{3n-1}}+{{2}^{3n-1}}+1 \\\end{align}$