Consider the following recurrence: $$f(1) = 1;$$ $$f(2n) = 2 f(n) - 1, \quad \text{for } n \ge 1;$$ $$f(2n+1) = 2 f(n) + 1, \quad \text{for } n \ge 1.$$ Then, which of the following statements is/are TRUE?

Let us compute values of $f(k)$ for small values of $k$ and test each statement:

• Base case: $f(1) = 1$

• For $n = 1$:

  • $f(2) = 2f(1) - 1 = 2(1) - 1 = 1$

  • $f(3) = 2f(1) + 1 = 2(1) + 1 = 3$

• For $n = 2$:

  • $f(4) = 2f(2) - 1 = 2(1) - 1 = 1$

  • $f(5) = 2f(2) + 1 = 2(1) + 1 = 3$

• For $n = 3$:

  • $f(6) = 2f(3) - 1 = 2(3) - 1 = 5$

  • $f(7) = 2f(3) + 1 = 2(3) + 1 = 7$

• For $n = 4$:

  • $f(8) = 2f(4) - 1 = 2(1) - 1 = 1$

  • $f(9) = 2f(4) + 1 = 2(1) + 1 = 3$

• For $n = 5$:

  • $f(10) = 2f(5) - 1 = 2(3) - 1 = 5$

Now, evaluate each option:

1. Option (A):$f(2^n - 1) = 2^n - 1$

   • For $n=1$: $f(1) = 1 = 2^1 - 1$.

   • For $n=2$: $f(3) = 3 = 2^2 - 1$.

   • By induction: $f(2^n - 1) = f(2(2^{n-1}-1) + 1) = 2f(2^{n-1}-1) + 1 = 2(2^{n-1}-1) + 1 = 2^n - 1$.

   • Option (A) is TRUE.

2. Option (B):$f(2^n) = 1$

   • For $n=1$: $f(2) = 1$.

   • By induction: $f(2^n) = 2f(2^{n-1}) - 1 = 2(1) - 1 = 1$.

   • Option (B) is TRUE.

3. Option (C):$f(5 \cdot 2^n) = 2^{n+1} + 1$

   • For $n=0$: $f(5) = 3 = 2^{0+1} + 1$.

   • For $n=1$: $f(10) = 2f(5) - 1 = 2(3) - 1 = 5 = 2^{1+1} + 1$.

   • By induction: $f(5 \cdot 2^n) = 2f(5 \cdot 2^{n-1}) - 1 = 2(2^n + 1) - 1 = 2^{n+1} + 1$.

   • Option (C) is TRUE.

4. Option (D):$f(2^n + 1) = 2^n + 1$

   • For $n=2$: $f(2^2 + 1) = f(5) = 3 \neq 2^2 + 1 = 5$.

   • Option (D) is FALSE.

Thus, statements A, B, and C are TRUE.