Với $n=1$: $7.2^{2.1-2}+3^{2.1-1}=7+3=10$ chia hết cho $5$.

Giả sử $7.2^{2k-2}+3^{2k-1}$ chia hết cho 5.
Ta sẽ chứng minh $7.2^{2(k+1)-2}+3^{2(k+1)-1}=7.2^{2k}+3^{2k+1}$ cũng chia hết cho $5$.

Thật vậy: $$\begin{aligned}7.2^{2k}+3^{2k+1}&=7.2^2\dfrac{2^{2k}}{2^2}+3^2\dfrac{3^{2k+1}}{3^2}\\
&=7.2^2.2^{2k-2}+3^2.3^{2k-1}\\
&=28.2^{2k-2}+9.3^{2k-1}\\
&=35.2^{2k-2}+10.3^{2k-1}-7.2^{2k-2}-3^{2k-1}\\
&=5\left(7.2^{2k-2}+2.3^{2k-1}\right)-\left(7.2^{2k-2}+3^{2k-1}\right)\;(*)\end{aligned}$$
Dễ thấy (*) chia hết cho $5$.

Vậy $7.2^{2n-2}+3^{2n-1}$ chia hết cho $5$, $\forall n\in\mathbb{N}^*$.