$\begin{align}  & \text{For }n=1,{{7}^{2(1)+1}}+1=344=43(8) \\ & \text{For }n=2,\text{           }{{7}^{2(2)+1}}+1=16808=2101(8) \\ & \text{Let assume }{{7}^{2n+1}}+1\text{ is divisible by 8 when }n=k \\ & {{7}^{2k+1}}+1=8M \\ & {{7}^{2k+1}}=8M-1---(i) \\ & \text{We need to show that }{{\text{7}}^{2n+1}}+1\text{ is divisible when }n=k+1 \\ & {{7}^{2(k+1)+1}}+1={{7}^{2k+3}}+1={{7}^{2}}({{7}^{2k+1}})+1 \\ & {{7}^{2(k+1)+1}}+1={{7}^{2}}(8M-1)+1 \\ & {{7}^{2(k+1)+1}}+1=49(8M-1)+1 \\ & {{7}^{2(k+1)+1}}+1=49(8M)-49+1 \\ & {{7}^{2(k+1)+1}}+1=49(8M)-48 \\ & {{7}^{2(k+1)+1}}+1=8(49M-6)---(ii) \\ & \text{Hence, }{{7}^{2n+1}}+1\text{ is divisible by }8 \\\end{align}$

\[\begin{align}  & \text{For }n=1,{{7}^{2(1)+1}}+1=344=43(8) \\ & \text{For }n=2,\text{           }{{7}^{2(2)+1}}+1=16808=2101(8) \\ & \text{Let assume }{{7}^{2n+1}}+1\text{ is divisible by 8 when }n=k \\ & {{7}^{2k+1}}+1=8M \\ & {{7}^{2k+1}}=8M-1---(i) \\ & \text{We need to show that }{{\text{7}}^{2n+1}}+1\text{ is divisible when }n=k+1 \\ & {{7}^{2(k+1)+1}}+1={{7}^{2k+3}}+1={{7}^{2}}({{7}^{2k+1}})+1 \\ & {{7}^{2(k+1)+1}}+1={{7}^{2}}(8M-1)+1 \\ & {{7}^{2(k+1)+1}}+1=49(8M-1)+1 \\ & {{7}^{2(k+1)+1}}+1=49(8M)-49+1 \\ & {{7}^{2(k+1)+1}}+1=49(8M)-48 \\ & {{7}^{2(k+1)+1}}+1=8(49M-6)---(ii) \\ & \text{Hence, }{{7}^{2n+1}}+1\text{ is divisible by }8 \\\end{align}\]