Question textWrite a function dseries(n_terms) to calculate the first n terms of the following seriesFor example:[table]
Test | Result print(dseries(3)) | 14
[/table] Answer:(penalty regime: 0, 10, 20, ... %)def dseries(n_terms): total=0 for i in range(1,n_terms+1): total+= i **2 return total

Question

Question textWrite a function dseries(n_terms) to calculate the first n terms of the following seriesFor example:[table]
Test | Result print(dseries(3)) | 14
[/table] Answer:(penalty regime: 0, 10, 20, ... %)def dseries(n_terms): total=0 for i in range(1,n_terms+1): total+= i **2 return total

BlackTom AI · Accepted Answer

To begin, note that the problem describes a sequence where each term is the square of its position, and we are asked to sum the first n_terms of these squares.

First, recognize the mathematical target: the sum of the first n terms of the series 1^2 + 2^2 + 3^2 + ... + n^2. This is the classic sum of squares.

Next, evaluate the provided code snippet:
- The function initializes total to 0, which is a correct accumulator.
- The loop uses for i in range(1, n_terms+1): which iterates i from 1 through n_terms inclusive. This matches the intended terms 1^2, 2^2, ..., n_terms^2.
- Inside the loop, total += i ** 2 adds the square of i to the total, so after the loop, total = 1^2 + 2^2 + ... + n_terms^2.
- Finally, it returns total, which yields the desired result for any positive n_terms.

Therefore, for the example dseries(3), the computation is 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14, matching the given test output.

Edge cases and small considerations:
- If n_terms is 0 or negative, the loop body does not execute (or the function may rely on calling code to avoid such inputs), and total remains 0, which is a reasonable result for a sum over an empty set.
- For large n_terms, Python integers handle big numbers, so there is no overflow, but performance scales linearly with n_terms due to the loop.

Alternative approach:
- If you want a closed-form formula, you can use the known formula for the sum of squares: n(n + 1)(2n + 1) / 6. Implementing this would reduce runtime to constant time, e.g., return n_terms * (n_terms + 1) * (2 * n_terms + 1) // 6.

In summary, the given implementation correctly computes the sum of squares for the first n_terms terms and aligns with the example provided (dseries(3) -> 14). If a more efficient constant-time solution is desired, the closed-form expression can be used instead.

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