CAT 2024 Slot 3 · Quant · Question 17

Only even-indexed terms appear in the sum, and the recurrence steps two at a time, so the even terms form a self-contained chain starting from {t}_{2}=-1.

Using {t}_{n}=\frac{n-3}{n-1}{t}_{n-2} with n=4, {t}_{4}=\frac{1}{3}{t}_{2}=-\frac{1}{3}.

With n=6, {t}_{6}=\frac{3}{5}{t}_{4}=-\frac{1}{5}.

With n=8, {t}_{8}=\frac{5}{7}{t}_{6}=-\frac{1}{7}.

The pattern is clear: writing n=2k, the term is {t}_{2k}=-\frac{1}{2k-1}.

Taking reciprocals, \frac{1}{{t}_{2k}}=-\left( 2k-1 \right), which is simply the negative of the k-th odd number.

The sum runs from {t}_{2} to {t}_{2024}, that is from k=1 to k=1012.

So the required sum is -\left( 1+3+5+\cdots +2023 \right).

The sum of the first m odd numbers is {m}^{2}, so this bracket equals {1012}^{2}.

Computing, {1012}^{2}=1024144.

The sum is -1024144, the third option.

Shortcut. Writing out three terms is enough to spot {t}_{2k}=-\frac{1}{2k-1}, after which the whole sum is the classic 1+3+5+\cdots ={m}^{2}. The answer must be a negative perfect square, and among the options only 1024144={1012}^{2} qualifies.