$\text{ If }n\text{ is a natural number, then }{4}^n-3n-1\text{ is divisible by which one of the following? }$

$\text{ Putting, }n=2,\text{ we get }{4}^n-3n-1=16-6-1=9$

$\text{ Putting, }n=3\text{ , we get }{4}^n-3n-1=64-9-1=54$

$\text{ Putting, }n=4,\text{ we get }{4}^n-3n-1=256-12-1=243$

$\text{ Hence, for any value of }n,4^n-3n-1\text{ is }$$\text{ always divisible by }9\text{ . }$ 

  $$\begin{gathered} second method: \\ {\left(1+3\right)}^n-3n-1=\left[ 1+C\left(n,1\right)3+C\left(n,2\right)3^2+C\left(n,3\right)3^3+.....+3^n\right]-3n-1 \end{gathered}$$

                              = $9\left[C\left(n,2\right)+C\left(n,3\right)3........\right] which is divisible by 9$