Question

Using binomial theorem, prove that $3^{2n+2}-8n-9$ is divisible by 64, $n\in N$.

Answer 1

$3^{2n+2}-8n-9=9^{n+1}-8n-9...\left(1\right)$
Consider
$$\begin{gathered} 9^{n+1}={\left(1+8\right)}^{n+1} \\ \Rightarrow 9^{n+1}={}^{n+1}C_0\times 8^0+{}^{n+1}C_1\times 8^1+{}^{n+1}C_2\times 8^2+{}^{n+1}C_3\times 8^3+...+{}^{n+1}C_{n+1}\times 8^{n+1} \end{gathered}$$
$$\begin{gathered} \Rightarrow 9^{n+1}=1+8(n+1)+[{}^{n+1}C_2\times 8^2+{}^{n+1}C_3\times 8^3+...+{}^{n+1}C_{n+1}\times 8^{n+1}] \\ \Rightarrow 9^{n+1}-8n-9=64({}^{n+1}C_2+{}^{n+1}C_3\times 8^1+...+{}^{n+1}C_{n+1}\times 8^{n-1}] \\ \Rightarrow 9^{n+1}-8n-9=64\times \mathrm{An}\mathrm{integer} \\ {9}^{n+1}-8n-9\mathrm{is}\mathrm{divisible}\mathrm{by}64 \\ \mathrm{Or}, \\ {3}^{2n+2}-8n-9\mathrm{is}\mathrm{divisible}\mathrm{by}64\left[\mathrm{From}\left(1\right)\right] \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$