The function ComputeSum receives a positive integer n as the input and returns the value

ComputeSum(n)
If n = 1, then
Return(3)
y := ComputeSum(n - 1)
Return( ? )

What is the correct value for the algorithm to return?

Question

The function ComputeSum receives a positive integer n as the input and returns the value

ComputeSum(n)
   If n = 1, then 
     Return(3)
   y := ComputeSum(n - 1)
     Return( ? )

What is the correct value for the algorithm to return?

BlackTom AI · Accepted Answer

Each recursive call should add the nth term (n + 2)·n^2 to the sum already computed for n − 1, so the procedure must return y + (n + 2)n^2.

1. Relevant concepts
- The function ComputeSum(n) is defined as the finite sum Σ_{j=1}^n (j + 2) j^2.
- A recursive implementation computes ComputeSum(n) by computing ComputeSum(n−1) and then adding the nth summand (n+2)n^2.
- Base case must match the sum for n = 1.

2. Step-by-step reasoning
- Base case: for n = 1 the summation gives (1+2)·1^2 = 3, so Return(3) is correct.
- Recursive step: assume y := ComputeSum(n−1) returns Σ_{j=1}^{n−1} (j+2) j^2.
- To obtain Σ_{j=1}^{n} (j+2) j^2 we add the nth term (n+2) n^2 to y. Thus ComputeSum(n) = y + (n+2) n^2.

3. Intermediate calculations (example)
- n = 1: ComputeSum(1) = (1+2)·1^2 = 3.
- n = 2: y = ComputeSum(1) = 3; add (2+2)·2^2 = 4·4 = 16; result = 3 + 16 = 19. Direct sum: (1+2)·1^2 + (2+2)·2^2 = 3 + 16 = 19, consistent.

4. Why the correct answer is correct
- The summation definition explicitly requires adding the nth summand to the previous partial sum. The term for index n is (n+2) n^2, so the recursive return must be y + (n+2) n^2.

5. Why the other choices are incorrect
- y: returns only the previous sum and never adds the nth term, so it undercounts.
- y+n: adds a linear term n, but the required nth summand grows like n^3 (since n^2 times (n+2) ≈ n^3), so this is wrong.
- (y+2)y^2: combines y with itself nonlinearly and does not represent adding the nth summand; wrong form.
- y+(n+2)n2 in the choices is intended as y+(n+2)n^2 and matches the correct recursive addition.

Conclusion: The correct return statement is y + (n + 2) n^2, which matches the provided answer y+(n+2)n^2.

The function ComputeSum receives a positive integer n as the input and returns the value ComputeSum(n) If n = 1, then Return(3) y := ComputeSum(n - 1) Return( ? ) What is the correct value for the algorithm to return? Single choice

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