# Show that

Question:

Show that $2^{4 n+4}-15 n-16$, where $n \in \mathbb{N}$ is divisible by 225 .

Solution:

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 .