Show that $2^{4 n+4}-15 n-16$, where $n \in \mathbb{N}$ is divisible by 225 .
We have:
$2^{4 n+4}-15 n-16=2^{4(n+1)}-15 n-16$
$=16^{n+1}-15 n-16$
$=(1+15)^{n+1}-15 n-16$
$={ }^{n+1} \mathrm{C}_{0} 15^{0}+{ }^{\mathrm{n}+1} \mathrm{C}_{1} 15^{1}+{ }^{\mathrm{n}+1} \mathrm{C}_{2} 15^{2}+\ldots+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1} 15^{\mathrm{n}+1}-15 \mathrm{n}-16$
$=1+(\mathrm{n}+1) 15+{ }^{\mathrm{n}+1} \mathrm{C}_{2} 15^{2}+\ldots+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1} 15^{\mathrm{n}+1}-15 \mathrm{n}-16$
$=1+15 n+15+{ }^{\mathrm{n}+1} \mathrm{C}_{2} 15^{2}+\ldots+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1} 15^{\mathrm{n}+1}-15 \mathrm{n}-16$
$={ }^{\mathrm{n}+1} \mathrm{C}_{2} 15^{2}+\ldots+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1} 15^{\mathrm{n}+1}$
$=15^{2}\left({ }^{\mathrm{n}+1} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1} 15^{\mathrm{n}-1}\right)$
$=225\left({ }^{\mathrm{n}+1} \mathrm{C}_{2}+\ldots+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1} 15^{\mathrm{n}-1}\right)$
Thus, $2^{4 n+4}-15 n-16$, where $n \in \mathbb{N}$ is divisible by 225 .