Consider the sequence t₁ = 1, t₂ = −1, and tₙ = [(n − 3)/(n − 1)]tₙ₋₂ for n ≥ 3. Then, the value of 1/t₂ + 1/t₄ + 1/t₆ + ⋯ + 1/t₂₀₂₄ + 1/t₂₀₂₆ is:
Answer and solution
Correct answer: B
Solution
Answer: B — −1,026,169
1. Focus on the even-numbered terms because the sum contains t₂, t₄, t₆ and so on.
2. For even n = 2k, the recurrence becomes:
3. t₂ₖ = [(2k − 3)/(2k − 1)] × t₂ₖ₋₂
4. We already know t₂ = −1.
5. Now calculate the next few terms:
6. t₄ = (1/3) × (−1) = −1/3
t₆ = (3/5) × (−1/3) = −1/5
t₈ = (5/7) × (−1/5) = −1/7
7. The pattern is clear:
8. t₂ = −1, t₄ = −1/3, t₆ = −1/5, t₈ = −1/7, and so on.
9. Taking reciprocals, the terms become:
10. −1, −3, −5, −7, ... , −2025
11. There are 1013 terms in this series.
12. The sum of the first n odd numbers is n². Therefore, the sum of these positive odd numbers is:
13. 1013² = 1,026,169
14. Since every term in our series is negative, the required sum is:
15. −1,026,169
16. Answer: −1,026,169.
Why this method works:
The recurrence connects terms two positions apart, so odd and even subsequences are independent. Only the even subsequence enters the sum. Its successive factors cancel: (1/3)(3/5)(5/7)…=1/(2k−1), with the initial negative sign retained.
Independent check and common trap:
The last even index 2026 corresponds to k=1013 and reciprocal−2025. The arithmetic-series check is 1013×(−1−2025)/2=−1013²=−1,026,169. Counting 2026 terms would double the intended number of terms.
Apply: t₂k=−1/(2k−1), so reciprocals are−1,−3,…,−2025. Count 1013; sum=−1013²=−1026169.
Check: The last even index 2026 corresponds to k=1013 and reciprocal−2025. The arithmetic-series check is 1013×(−1−2025)/2=−1013²=−1,026,169. Counting 2026 terms would double the intended number of terms.
1. Focus on the even-numbered terms because the sum contains t₂, t₄, t₆ and so on.
2. For even n = 2k, the recurrence becomes:
3. t₂ₖ = [(2k − 3)/(2k − 1)] × t₂ₖ₋₂
4. We already know t₂ = −1.
5. Now calculate the next few terms:
6. t₄ = (1/3) × (−1) = −1/3
t₆ = (3/5) × (−1/3) = −1/5
t₈ = (5/7) × (−1/5) = −1/7
7. The pattern is clear:
8. t₂ = −1, t₄ = −1/3, t₆ = −1/5, t₈ = −1/7, and so on.
9. Taking reciprocals, the terms become:
10. −1, −3, −5, −7, ... , −2025
11. There are 1013 terms in this series.
12. The sum of the first n odd numbers is n². Therefore, the sum of these positive odd numbers is:
13. 1013² = 1,026,169
14. Since every term in our series is negative, the required sum is:
15. −1,026,169
16. Answer: −1,026,169.
Why this method works:
The recurrence connects terms two positions apart, so odd and even subsequences are independent. Only the even subsequence enters the sum. Its successive factors cancel: (1/3)(3/5)(5/7)…=1/(2k−1), with the initial negative sign retained.
Independent check and common trap:
The last even index 2026 corresponds to k=1013 and reciprocal−2025. The arithmetic-series check is 1013×(−1−2025)/2=−1013²=−1,026,169. Counting 2026 terms would double the intended number of terms.
G Strategy shortcut
Spot: Algebra – Sequences and Recurrence.Apply: t₂k=−1/(2k−1), so reciprocals are−1,−3,…,−2025. Count 1013; sum=−1013²=−1026169.
Check: The last even index 2026 corresponds to k=1013 and reciprocal−2025. The arithmetic-series check is 1013×(−1−2025)/2=−1013²=−1,026,169. Counting 2026 terms would double the intended number of terms.
G Strategy: Spot: Algebra – Sequences and Recurrence. Apply: t₂k=−1/(2k−1), so reciprocals are−1,−3,…,−2025. Count 1013; sum=−1013²=−1026169. Check: The last even index 2026 corresponds to k=1013 and reciprocal−2025. The arithmetic-series check is 1013×(−1−2025)/2=−1013²=−1,026,169. Counting 2026 terms would double the intended number of terms.