Question

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

Answer 1

We have,
$$\begin{gathered} 2^{4n+4}-15n-16=2^{4\left(n+1\right)}-15n-16 \\ ={16}^{n+1}-15n-16 \\ ={\left(1+15\right)}^{n+1}-15n-16 \\ {=}^{n+1}{\mathrm{C}}_0{15}^0{+}^{\mathrm{n}+1}{\mathrm{C}}_1{15}^1{+}^{\mathrm{n}+1}{\mathrm{C}}_2{15}^2+...{+}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}{15}^{\mathrm{n}+1}-15\mathrm{n}-16 \\ =1+(\mathrm{n}+1)15{+}^{\mathrm{n}+1}{\mathrm{C}}_2{15}^2+...{+}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}{15}^{\mathrm{n}+1}-15\mathrm{n}-16 \\ =1+15\mathrm{n}+15{+}^{\mathrm{n}+1}{\mathrm{C}}_2{15}^2+...{+}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}{15}^{\mathrm{n}+1}-15\mathrm{n}-16 \\ {=}^{\mathrm{n}+1}{\mathrm{C}}_2{15}^2+...{+}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}{15}^{\mathrm{n}+1} \\ ={15}^2\left({}^{\mathrm{n}+1}{\mathrm{C}}_2+...{+}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}{15}^{\mathrm{n}-1}\right) \\ =225\left({}^{\mathrm{n}+1}{\mathrm{C}}_2+...{+}^{\mathrm{n}+1}{\mathrm{C}}_{\mathrm{n}+1}{15}^{\mathrm{n}-1}\right) \end{gathered}$$

Thus, ​ $2^{4n+4}-15n-16$, where n ∈ $\mathrm{\mathbb{N}}$ is divisible by 225.