CAT 2024 Slot 3 QA Question 18: Consider the sequence $t_1 = 1,\;t_2 = -1$ and $t_n = \left(\dfrac{n - 3}{n - 1}\right)t_{n - 2}$ for $n\geq 3$. Then, the value of the sum $\dfrac{1}{t_2} + \dfrac{1}{t_4} + \dfrac{1}{t_6} + \cdots +

Question

Consider the sequence $t_1 = 1,\;t_2 = -1$ and $t_n = \left(\dfrac{n - 3}{n - 1}\right)t_{n - 2}$ for $n\geq 3$. Then, the value of the sum $\dfrac{1}{t_2} + \dfrac{1}{t_4} + \dfrac{1}{t_6} + \cdots +

Accepted Answer

Only even-indexed terms are needed. Put $n = 2k$.

$t_{2k} = (\dfrac{2k - 3}{2k - 1})t_{2k - 2}$

Starting with $t_2 = -1$:

$t_4 = \dfrac{1}{3}(-1) = -\dfrac{1}{3}$
$t_6 = \dfrac{3}{5}\left(-\dfrac{1}{3}\right) = -\dfrac{1}{5}$

So the pattern is:

$t_{2k} = -\dfrac{1}{2k - 1}$
$\dfrac{1}{t_{2k}} =