Consider the sequence t1=1,t2=−1 and tₙ = [(n − 3) / (n − 1)] × tₙ₋₂ for n ≥ 3 Then, the value of the sum 1/t₂ + 1/t₄ + 1/t₆ + 1/t₈ + … + 1/t₂₀₂₄ is
Question:
Consider the sequence t1=1,t2=−1 and tₙ = [(n − 3) / (n − 1)] × tₙ₋₂ for n ≥ 3 Then, the value of the sum 1/t₂ + 1/t₄ + 1/t₆ + 1/t₈ + … + 1/t₂₀₂₄ is
Consider the sequence t1=1,t2=−1 and tₙ = [(n − 3) / (n − 1)] × tₙ₋₂ for n ≥ 3 Then, the value of the sum 1/t₂ + 1/t₄ + 1/t₆ + 1/t₈ + … + 1/t₂₀₂₄ is
Options
Answer: -1024144
Explanation:
We are asked to compute: 1/t₂ + 1/t₄ + 1/t₆ + 1/t₈ + … + 1/t₂₀₂₄ From the sequence pattern, the reciprocals of the even terms are: 1/t₂ = −1, 1/t₄ = −3, 1/t₆ = −5, 1/t₈ = −7, … , 1/t₂₀₂₄ = −2023 So the sum becomes: −(1 + 3 + 5 + 7 + … + 2023) There are 1012 odd numbers from 1 to 2023. Sum of first 1012 odd numbers = (1012)² = 1,024,144 Hence, the total sum = −1,024,144
Explanation:
We are asked to compute: 1/t₂ + 1/t₄ + 1/t₆ + 1/t₈ + … + 1/t₂₀₂₄ From the sequence pattern, the reciprocals of the even terms are: 1/t₂ = −1, 1/t₄ = −3, 1/t₆ = −5, 1/t₈ = −7, … , 1/t₂₀₂₄ = −2023 So the sum becomes: −(1 + 3 + 5 + 7 + … + 2023) There are 1012 odd numbers from 1 to 2023. Sum of first 1012 odd numbers = (1012)² = 1,024,144 Hence, the total sum = −1,024,144
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