Consider the sequence $t_1 = 1$, $t_2 = -1$ and $t_n = \left(\frac{n-3}{n-1}\right) t_{n-2}$ for $n \ge 3$. The, the value of the sum $\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \dots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}$ is

Most questions like this require you to find a pattern of the sequence.

We have a sequence defined by:

$t_1 = 1$

$t_2 = -1$

$t_n = \left(\frac{n-3}{n-1}\right) t_{n-2}$ for $n \geq 3$ (recursive formula)

Each term depends on the term that came two positions before it. This is called a "second-order" recurrence relation.

Since we need $\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \ldots$, we only need to find the even-indexed terms.

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Let's calculate the first few even-indexed terms:

For $t_4$ (using $n = 4$):

$t_4 = \left(\frac{4-3}{4-1}\right) \times t_2$

$= \frac{1}{3} \times (-1)$

$= -\frac{1}{3}$

For $t_6$ (using $n = 6$):

$t_6 = \left(\frac{6-3}{6-1}\right) \times t_4$

$= \frac{3}{5} \times \left(-\frac{1}{3}\right)$

$= -\frac{1}{5}$

For $t_8$ (using $n = 8$):

$t_8 = \left(\frac{8-3}{8-1}\right) \times t_6$

$= \frac{5}{7} \times \left(-\frac{1}{5}\right)$

$= -\frac{1}{7}$

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Notice the pattern:

$t_2 = -1 = -\frac{1}{1}$

$t_4 = -\frac{1}{3}$

$t_6 = -\frac{1}{5}$

$t_8 = -\frac{1}{7}$

The denominators are consecutive odd numbers: 1, 3, 5, 7, ...

Pattern: $t_{2k} = -\frac{1}{2k-1}$ for any positive integer $k$.

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Question asks for the sum of the reciprocal.

Since $t_{2k} = -\frac{1}{2k-1}$, we have:

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

$= -(2k-1)$

The reciprocal of each even-indexed term is the negative of an odd number.

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We need: $\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \ldots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}$

Using our formula $\frac{1}{t_{2k}} = -(2k-1)$:

$\frac{1}{t_2} = -(2 \times 1 - 1) = -1$

$\frac{1}{t_4} = -(2 \times 2 - 1) = -3$

$\frac{1}{t_6} = -(2 \times 3 - 1) = -5$

$\dots$

$\frac{1}{t_{2022}} = -(2 \times 1011 - 1) = -2021$

$\frac{1}{t_{2024}} = -(2 \times 1012 - 1) = -2023$

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Our sum becomes:

$(-1) + (-3) + (-5) + \ldots + (-2021) + (-2023)$

$= -(1 + 3 + 5 + \ldots + 2021 + 2023)$

This requires the sum of odd numbers from $1$ to $2023$

Since, $2023$ is odd, number of odd terms = $\frac{2023 + 1}{2} = 1012$

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The sum of the first $n$ odd natural numbers equals $n^2$.

$-(1 + 3 + 5 + \ldots + 2023) = -(1012^2) = -1024144$

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Hence:

$\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \ldots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}} = -1024144$