Question

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

Answer 1

$2^{3n}-7n-1$

$=2^{3\left(n\right)}-7\left(n\right)-1$

$=8^n-7n-1$

$=(1+7{)}^m-7n-1$

$={(}^n{C}_0{+}^n{C}_1\left(7{)}^1{+}^n{C}_2\right(7{)}^2+...{.}^n{C}_n\left(7{)}^n\right)-7n-1$

$=(1-7n+{49}^n{C}_2+....+49(7{)}^{n-2})-7n-1$

$=49{(}^n{C}_2+....+7^{n-2})$

$\therefore 2^{3n}-7n-1$ is divisible by 49 Hence, proved.