Chapter 6: Problem 19
Consider the following recursive method JPS private static int JPS(int n) { if(n == 0) return 1; return 3*JPS(n-1)+1; } Which corresponds to the recursive series: JPS(0) = 1 JPS(n) = 3.JPS(n-1)+1 Write a driver that prints out the first 10 values of the series. What is the closed form of JPS?
Short Answer
Step by step solution
Evaluate the Base Case
Recursive Evaluation up to JPS(9)
Identify the Pattern
Derive the Closed Form
Verify the Closed Form
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Recurrence Relation
- Base Case: \( JPS(0) = 1 \)
- Recursive Case: \( JPS(n) = 3 \times JPS(n-1) + 1 \)
Geometric Progression
- The multiplier in the recursion is 3, acting as the geometric ratio.
- Every term is constructed using the previous term multiplied by 3, with an added 1, suggesting a modified progression.
Closed Form Solution
Driver Program
- Initiating a loop to iterate from 0 to 9.
- Calling the JPS function within the loop for each \( n \).
- Printing the results to validate that the recursive and closed form produce consistent outputs.