Question:

/**
 * This method returns the sum of the first n numbers in the series defined by the recurrence relation:
 * a(1) = 2
 * a(n) = a(n-1) + 3, for n > 1
 * 
 * Complete the recursive method below to implement this functionality.
 */

public static int recursiveSum(int n) {
    // TODO: Implement the recursive sum method
    
    int sum;
    
    if (n == 1) {
        return 2;
    } else {
        sum = recursiveSum(n - 1) + 3;
    }
    
    return sum;
}

Answer:

To find the sum of the first n numbers in the given series, we can use a recursive approach. The recursive function recursiveSum() is implemented as follows:

public static int recursiveSum(int n) {
    int sum;
    
    if (n == 1) {
        // Base case: if n is 1, return the initial value of the series, which is 2
        return 2;
    } else {
        // Recursive case: calculate the sum of the previous term and 3
        sum = recursiveSum(n - 1) + 3;
    }
    
    return sum;
}

Explanation:

The base case is when n becomes 1. In this case, we have reached the first term of the series, which is 2. Therefore, we simply return 2.

For the recursive case (when n is greater than 1), we need to calculate the sum of the current term (a(n)) and the sum of the previous terms (a(n-1)). According to the recurrence relation, the next term (a(n)) is obtained by adding 3 to the previous term (a(n-1)).

To calculate the sum of the previous terms, we recursively call the recursiveSum() method with n-1, which reduces the problem to finding the sum of a smaller series. This continues recursively until we reach the base case.

Once the final recursive call returns, we add 3 to the sum of the previous terms to obtain the sum of the current term. This process continues until we reach the initial term (a(1)).

Example:

Let's call the recursiveSum() method with n = 4. We will trace the recursive calls step by step:

recursiveSum(4)
    recursiveSum(3)
        recursiveSum(2)
            recursiveSum(1) -> Base case reached, returns 2
        sum = recursiveSum(2) + 3 = 2 + 3 = 5
    sum = recursiveSum(3) + 3 = 5 + 3 = 8
sum = recursiveSum(4) + 3 = 8 + 3 = 11

Thus, the sum of the first 4 numbers in the series is 11.