Question

Using binomial theorem, prove that $2^{3n}-7n-1$ is divisible by 49, where $n\in N$.

Answer 1

$2^{3n}-7n-1=8^n-7n-1$ ...(1)

$$\begin{gathered} \mathrm{Now}, \\ {8}^n=(1+7{)}^n \\ ={}^{n}C_0+{}^{n}C_1\times 7^1+{}^{n}C_2\times 7^2+{}^{n}C_3\times 7^3+{}^{n}C_4\times 7^4+...+{}^{n}C_n\times 7^n \\ \Rightarrow 8^n=1+7n+49[{}^{n}C_2+{}^{n}C_3\times 7^1+{}^{n}C_4\times 7^2+...+{}^{n}C_n\times 7^{n-2}] \\ \Rightarrow 8^n-1-7n=49\times \left(\mathrm{An}\mathrm{integer}\right) \\ \mathrm{Now}, \\ {8}^n-1-7n\mathrm{is}\mathrm{divisible}\mathrm{by}49 \\ \mathrm{Or}, \\ {2}^{3n}-1-7n\mathrm{is}\mathrm{divisible}\mathrm{by}49\left[\mathrm{From}\left(1\right)\right] \end{gathered}$$