Sum of reciprocal terms in recurrence sequence

CAT 2024 Slot 3 · QA · Hard · Algebra

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Consider the sequence t1=1,t2=−1t_1 = 1, t_2 = -1 and tn=(n−3n−1)tn−2t_n = \left(\frac{n-3}{n-1}\right) t_{n-2} for n≥3n \ge 3. Then, the value of the sum 1t2+1t4+1t6+⋯+1t2022+1t2024\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \dots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}, is

  1. A.

    -1023132

  2. B.

    -1022121

  3. C.

    -1024144

  4. D.

    -1026169

Answer

C

Explanation

We evaluate terms with even indices:

  • t2=−1t_2 = -1
  • t4=13t2=−13t_4 = \frac{1}{3} t_2 = -\frac{1}{3}
  • t6=35t4=−15t_6 = \frac{3}{5} t_4 = -\frac{1}{5}
  • In general, t2k=−12k−1t_{2k} = -\frac{1}{2k - 1}, which implies 1t2k=−(2k−1)\frac{1}{t_{2k}} = -(2k - 1).

The required sum is: ∑k=110121t2k=−∑k=11012(2k−1)\sum_{k=1}^{1012} \frac{1}{t_{2k}} = -\sum_{k=1}^{1012} (2k - 1)

The sum of the first NN odd positive integers is N2N^2. Here N=1012N = 1012: Sum=−(1012)2=−(1000+12)2=−(1000000+24000+144)=−1024144\text{Sum} = -(1012)^2 = -(1000 + 12)^2 = -(1000000 + 24000 + 144) = -1024144

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