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For natural numbers $n$, the inequality $2^n >2n+1$ is valid when

  1. $n \geq 3$
  2. $n < 3$
  3. $n=3$
  4. None of these
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We know that $2^n$ would always be greater than $2*n$(exponential increase > linear increase) after a certain point.

We just need to find that point.

The intersection point would occur when

$2^n = 2n+1$

On solving this we get $n \approx 2.6$.

Now after this i.e. after $n=2.6$ the inequality would hold for all values of $n$

$\therefore$ Option $A.$ is correct answer.

 

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