Consider the sequence $$ t_1 = 1 $$, $$ t_2 = -1 $$, and
$$
t_n = \left(\frac{n-3}{n-1} ight) t_{n-2} \quad ext{for } n \geq 3.
$$
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:

Question

Consider the sequence $$ t_1 = 1 $$, $$ t_2 = -1 $$, and  
$$
t_n = \left(\frac{n-3}{n-1}ight) t_{n-2} \quad 	ext{for } n \geq 3.
$$  
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:

Accepted Answer

1. **Define the sequence terms recursively:**  
   - Starting with $$ t_1 = 1 $$, $$ t_2 = -1 $$:  
     For $$ n \geq 3 $$:  
     $$
     t_n = \left(\frac{n-3}{n-1}ight) t_{n-2}.
     $$

2. **Calculate specific terms in the sequence:**  
   - $$ t_2 = -1 $$.  
   - $$ t_4 = \left(\frac{4-3}{4-1}ight) t_2 = \frac{1}{3} 	imes (-1) = -\frac{1}{3} $$.  
   - $$ t_6 = \left(\frac{6-3}{6-1}ight) t_4 = \frac{3}{5} 	imes (-\frac{1}{3}) = -\frac{1}{5} $$.  
   - $$ t_8 = \left(\frac{8-3}{8-1}ight) t_6 = \frac{5}{7} 	imes (-\frac{1}{5}) = -\frac{1}{7} $$.

Continuing this pattern, we observe that:  
   - $$ t_{2k} = -\frac{1}{2k-1} $$ for even indices $$ n = 2k $$.

3. **Sum of reciprocals for even indices:**  
   The sum becomes:  
   $$
   S = \frac{1}{t_2} + \frac{1}{t_4} + \cdots + \frac{1}{t_{2024}}.
   $$  
   Substituting $$ t_{2k} = -\frac{1}{2k-1} $$:  
   $$
   S = -(1 + 3 + 5 + 7 + \cdots + 2023).
   $$

4. **Sum of first 1012 odd numbers:**  
   The sum of the first $$ n $$ odd numbers is given by:  
   $$
   n^2.
   $$  
   Here, there are 1012 odd numbers (from 1 to 2023):  
   $$
   S = -(1012^2).
   $$

5. **Calculate the final value:**  
   Using $$ 1012^2 = 1024144 $$:  
   $$
   S = -1024144.
   $$

Thus, the value of the sum is **-1024144**.

Answer

Answer

Answer