For $n \in \mathbb{N}$, let

$f(n)= \begin{cases}3 n+1 & \text { if } n \text { is odd } \\ \frac{n}{2} & \text { if } n \text { is even }\end{cases}$

Define $f^{(1)}(n)=f(n)$ and $f^{(i)}(n)=f\left(f^{(i-1)}(n)\right)$ for $i \gt 1$. Suppose $l$ is an even number greater than or equal to 2020. Then the least value of $i$ for which

$f^{(i)}\left(\frac{2^{\ell}\left(2^{\ell}-1\right)}{3}\right)=1$
is

• A

  $2 \ell-1$
• B

  $2 \ell+2$
• Correct answer
  C

  $2 \ell+1$
• D

  $2 \ell$

Hence, the correct answer is option: 'C'.