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CAT 2024 Set-3 | Quantitative Aptitude | Question: 3

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Consider the sequence $t_{1}=1, t_{2}=-1$ and $t_{n}=\left(\frac{n-3}{n-1}\right) t_{n-2}$ for $n \geq 3$. Then, the value of the sum $\frac{1}{t_{2}}+\frac{1}{t_{4}}+\frac{1}{t_{6}}+\cdots+\frac{1}{t_{2022}}+\frac{1}{t_{2024}}$, is

  1. $-1024144$
  2. $-1026169$
  3. $-1022121$
  4. $-1023132$

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