Question

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

• A. -1024144
• B. -1023132
• C. -1026169
• D. -1022121

Answer 1

The Correct Option is A

We are given the recurrence relation $t_n = \frac{n-3}{n-1} t_{n-2}$ for $n \ge 3$, with initial terms $t_1 = 1$ and $t_2 = -1$.
We need to compute the sum:
\[ S = \frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \dots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}} \]

Let's find the first few terms using the recurrence:
Nbsp;\(t_3\)\(= \frac{3-3}{3-1} t_1 = 0\)Nbsp;

\(t_4 = \frac{4-3}{4-1} t_2 = \frac{1}{3} \times (-1) = -\frac{1}{3}\)Nbsp;Nbsp;

\(t_5 = \frac{5-3}{5-1} t_3 = \frac{2}{4} \times 0 = 0\)Nbsp;

\(t_6 = \frac{6-3}{6-1} t_4 = \frac{3}{5}\)\(\times \left(-\frac{1}{3}\right) = -\frac{1}{5}\)

We observe a pattern for terms with even indices: $t_2 = -1$, $t_4 = -\frac{1}{3}$, $t_6 = -\frac{1}{5}$, and so on. For even $n$, $t_n = -\frac{1}{n-1}$.

Now we can rewrite the sum as:
\[ S = \sum_{k=1}^{1012} \frac{1}{t_{2k}} = \sum_{k=1}^{1012} \frac{1}{-\frac{1}{(2k)-1}} = \sum_{k=1}^{1012} -(2k-1) = -\sum_{k=1}^{1012} (2k-1) \]

The sum of the first 1012 odd numbers is $1012^2$. Therefore:
\[ S = -1012^2 = -1024144 \]