Answer A: Google Gemini 2.5 Flash-Lite
1. Proof that every term of the sequence is positive: We can prove this by induction. Base case: a1 = 1, which is positive. Inductive step: Assume that ak is positive for some integer k >= 1. We need to show that ak+1 is also positive. ak+1 = (ak + 2/ak) / 2. Since ak > 0, then 2/ak > 0. The sum of two positive numbers (ak and 2/ak) is positive. Dividing a positive number by 2 results in a positive number. Therefore, ak+1 > 0. By induction, every term of the sequence is positive. 2. Proof that the sequence is decreasing after the first term: We need to show that an+1 <= an for n >= 1. an+1 - an = (an + 2/an...
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1. Proof that every term of the sequence is positive: We can prove this by induction. Base case: a1 = 1, which is positive. Inductive step: Assume that ak is positive for some integer k >= 1. We need to show that ak+1 is also positive. ak+1 = (ak + 2/ak) / 2. Since ak > 0, then 2/ak > 0. The sum of two positive numbers (ak and 2/ak) is positive. Dividing a positive number by 2 results in a positive number. Therefore, ak+1 > 0. By induction, every term of the sequence is positive. 2. Proof that the sequence is decreasing after the first term: We need to show that an+1 <= an for n >= 1. an+1 - an = (an + 2/an) / 2 - an = (an + 2/an - 2an) / 2 = (2/an - an) / 2 = (2 - an^2) / (2an). For the sequence to be decreasing, we need an+1 - an <= 0, which means (2 - an^2) / (2an) <= 0. Since we have already proven that all terms are positive (an > 0), the denominator 2an is positive. Therefore, we need the numerator to be non-positive: 2 - an^2 <= 0, which implies an^2 >= 2. Let's check if an^2 >= 2 for n >= 2. For n=1, a1 = 1, a1^2 = 1, which is not >= 2. So the sequence is not decreasing from the first term. For n=2, a2 = (a1 + 2/a1) / 2 = (1 + 2/1) / 2 = 3/2. a2^2 = (3/2)^2 = 9/4 = 2.25, which is >= 2. Now, let's prove by induction that an^2 >= 2 for n >= 2. Base case: We've shown a2^2 = 9/4 >= 2. Inductive step: Assume ak^2 >= 2 for some integer k >= 2. We need to show that ak+1^2 >= 2. ak+1 = (ak + 2/ak) / 2. ak+1^2 = ((ak + 2/ak) / 2)^2 = (ak^2 + 4 + 4/ak^2) / 4. Since ak^2 >= 2, then 1/ak^2 <= 1/2. So, ak+1^2 = (ak^2 + 4 + 4/ak^2) / 4 >= (2 + 4 + 4(0)) / 4 = 6/4 = 1.5. This is not sufficient. Let's use a different approach for the inductive step for an^2 >= 2 for n >= 2. We know ak+1 = (ak + 2/ak) / 2. Consider ak+1^2 - 2 = ((ak + 2/ak) / 2)^2 - 2 = (ak^2 + 4 + 4/ak^2) / 4 - 2 = (ak^2 + 4 + 4/ak^2 - 8) / 4 = (ak^2 - 4 + 4/ak^2) / 4 = (ak - 2/ak)^2 / 4. Since (ak - 2/ak)^2 is a square, it is always non-negative. Thus, ak+1^2 - 2 >= 0, which means ak+1^2 >= 2. Therefore, for n >= 2, an^2 >= 2, which implies that an+1 - an <= 0. The sequence is decreasing after the first term. 3. Proof that the sequence is bounded below: From part 1, we have proven that every term of the sequence is positive. Thus, the sequence is bounded below by 0. Furthermore, from part 2, we showed that for n >= 2, an^2 >= 2, which implies an >= sqrt(2) for n >= 2. Therefore, the sequence is bounded below by sqrt(2) for n >= 2. 4. Deduction that the sequence converges: From part 2, we have shown that the sequence is decreasing for n >= 2. From part 3, we have shown that the sequence is bounded below (by sqrt(2) for n >= 2). According to the Monotone Convergence Theorem, a sequence that is both monotone (decreasing in this case) and bounded below converges to a finite limit. 5. Finding the limit and justification: Let L be the limit of the sequence. Since the sequence converges, we can take the limit of both sides of the recurrence relation: lim (an+1) = lim ((an + 2/an) / 2) As n approaches infinity, an+1 approaches L and an approaches L. So, L = (L + 2/L) / 2. Now, we solve for L: 2L = L + 2/L L = 2/L L^2 = 2 L = ±sqrt(2). Since we have proven that all terms of the sequence are positive (part 1), the limit must also be positive. Therefore, L = sqrt(2). Justification: The limit is sqrt(2) because the sequence is proven to be decreasing and bounded below by sqrt(2) for n >= 2, and the limit of the recurrence relation must satisfy L = (L + 2/L) / 2, which yields L = sqrt(2) as the only positive solution.
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Overall Comments
Answer A provides a complete and correct solution to all parts of the problem. It successfully uses induction to prove the positivity of the terms and a separate, more complex inductive proof to establish that the sequence is decreasing from the second term onward. While the logic is sound, the presentation of the proof for monotonicity is slightly convoluted, as it includes a failed attempt before presenting a successful one. The justifications for convergence and the limit are correct but lack the finer points of rigor seen in the alternative answer.
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Correctness
Weight 45%The answer is entirely correct. All proofs are valid, and the final limit is correct. The self-correction during the proof of monotonicity is handled well and leads to a correct argument.
Reasoning Quality
Weight 20%The reasoning is logical and sound. The proof that an^2 >= 2 for n >= 2 is correct. However, the method is less elegant than the alternative, and the inclusion of a failed attempt, while honest, makes the argument less direct.
Completeness
Weight 15%The answer addresses all five parts of the prompt thoroughly and provides a full solution.
Clarity
Weight 10%The answer is clearly structured, following the numbering of the prompt. The writing is easy to follow, though the section on monotonicity is slightly convoluted due to the presentation of a failed attempt before the successful proof.
Instruction Following
Weight 10%The answer perfectly follows the instructions, answering all parts of the prompt in the style of a rigorous exam solution.
Total Score
Overall Comments
Positivity and the limit computation are handled correctly, and it correctly notes the sequence increases from a1 to a2 then decreases afterward. However, the monotonicity proof is messy: it starts an induction for a_n^2≥2 but makes an incorrect inequality step, then switches approaches. Although it eventually lands on a correct identity implying a_{k+1}^2≥2, the exposition has a noticeable logical stumble. The bounded-below part is also a bit confused: it first gives 0 as a lower bound, then claims a_n≥√2 for n≥2 without clearly emphasizing that this stronger bound is what supports the monotone convergence argument for the tail. Overall correct but with weaker rigor and presentation issues.
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Correctness
Weight 45%Final limit √2 is correct and the core claims are mostly established, but the monotonicity section includes an incorrect inequality step and a somewhat patched argument, reducing reliability.
Reasoning Quality
Weight 20%Reasoning is partially muddled: it attempts an induction with a wrong bound before switching methods; the final argument works but the logical flow is not consistently tight.
Completeness
Weight 15%Addresses all five parts, but parts 2–3 are less cleanly connected (lower bound and monotonicity dependence could be stated more crisply).
Clarity
Weight 10%Understandable but includes a false start and extra algebra that obscures the key ideas.
Instruction Following
Weight 10%Follows the exam-solution style and answers all parts, though the rigor in monotonicity is a bit uneven.
Total Score
Overall Comments
Answer A provides a correct and complete solution to all five parts of the problem. The positivity proof by induction is clear and correct. The monotonicity proof correctly identifies that an²≥2 for n≥2 is needed and proves it, though the initial failed attempt (using a weaker bound) before arriving at the correct algebraic identity (an+1² - 2 = (an - 2/an)²/4) is somewhat messy. The boundedness, convergence, and limit computation are all correct. However, the presentation suffers from the false start in part 2, which makes the proof less elegant and slightly harder to follow.
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Correctness
Weight 45%All mathematical statements and proofs are correct. The key identity an+1² - 2 = (an - 2/an)²/4 is correctly derived and used. The limit computation is correct. Minor issue: the initial failed attempt in part 2 doesn't affect correctness but shows some inefficiency in the proof strategy.
Reasoning Quality
Weight 20%The reasoning is ultimately sound but the false start in Part 2 (attempting a bound that doesn't work before finding the correct approach) detracts from the quality. The final algebraic identity approach works but is less insightful than the AM-GM approach.
Completeness
Weight 15%All five parts are addressed with complete proofs. The answer establishes positivity, monotonicity, boundedness, convergence, and the correct limit. Could have been more explicit about why passing to the limit in the recurrence is valid.
Clarity
Weight 10%The presentation is generally clear but the false start in Part 2 (trying a bound that doesn't work, then switching approaches) makes the proof harder to follow. The writing is functional but not polished.
Instruction Following
Weight 10%Follows the instructions by addressing all five parts and writing in a proof-style format. The false start in Part 2 is somewhat inappropriate for a 'rigorous exam solution' as requested.