1. Base inductiva ($n=0$):
$$ 2^5 \cdot 3^0 + 5^1 = 32 \cdot 1 + 5 = 37 $$
Verificado, $37$ es divisible por $37$.

2. Hipótesis:
$2^{k+5} \cdot 3^{4k} + 5^{3k+1} = 37m$.

3. Paso inductivo para $n=k+1$:
$$ \begin{aligned} E &= 2^{k+1+5} \cdot 3^{4(k+1)} + 5^{3(k+1)+1} \\ &= 2^1 \cdot 3^4 (2^{k+5} \cdot 3^{4k}) + 5^3 \cdot 5^{3k+1} \\ &= 2 \cdot 81 (2^{k+5} \cdot 3^{4k}) + 125 \cdot 5^{3k+1} \\ &= 162 (2^{k+5} \cdot 3^{4k}) + 125 \cdot 5^{3k+1} \end{aligned} $$
Expresamos $162$ en términos de $125$: $162 = 125 + 37$.
$$ \begin{aligned} E &= (125 + 37)(2^{k+5} \cdot 3^{4k}) + 125 \cdot 5^{3k+1} \\ &= 37(2^{k+5} \cdot 3^{4k}) + 125(2^{k+5} \cdot 3^{4k} + 5^{3k+1}) \end{aligned} $$
Usando la hipótesis inductiva:
$$ \begin{aligned} E &= 37(2^{k+5} \cdot 3^{4k}) + 125(37m) \\ &= 37(2^{k+5} \cdot 3^{4k} + 125m) \end{aligned} $$

$$ \boxed{(2^{n+5} \times 3^{4n} + 5^{3n+1}) \text{ es divisible por } 37} $$