Question

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

• A. \( f(2n - 1) = 2n - 1 \)
• B. \( f(2^n) = 1 \)
• C. \( f(5 \cdot 2^n) = 2^{n+1} + 1 \)
• D. \( f(2^n + 1) = 2^n + 1 \)

Hint:

For recurrence relations, carefully handle even and odd cases. The recurrence formula simplifies when working with powers of 2 or similar recursive structures.

Answer 1

The Correct Option is A, B, C

We are given a recurrence relation with specific rules for even and odd values of \( n \). We can use these rules to compute the values for different expressions: Calculation of \( f(2n-1) \): From the recurrence relation: \[ f(2n + 1) = 2f(n) + 1 \] Using this, we can directly compute \( f(2n-1) = 2n-1 \). This matches Option (A). Calculation of \( f(2^n) \): The recurrence relation tells us that for even \( n \), \( f(2n) = 2f(n) - 1 \). This can be iteratively calculated to get the value of \( f(2^n) \). However, by following the given rule and computing values for small \( n \), we observe that: \[ f(2^n) = 1 \] So, Option (B) is correct. Calculation of \( f(5 \cdot 2^n) \): For this case, it can be derived that: \[ f(5 \cdot 2^n) = 2^{n+1} + 1 \] So, Option (C) is correct. Final Verification of \( f(2^n + 1) \): For odd \( n \), we find that: \[ f(2^n + 1) = 2^n + 1 \] Thus, Option (D) is incorrect.
Thus, the correct answers are (A), (B), and (C).
Final Answer: (A), (B), (C)