\( a \) and \( b \) are natural numbers, then the value of \( a + b \) is: \[ (a + b\sqrt{3})^2 = 52 + 30\sqrt{3} \]

✅ Solution:

Expanding the left-hand side: \[ (a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2 \] Match with the right-hand side: \[ a^2 + 3b^2 = 52 \quad \text{(1)} \\ 2ab = 30 \Rightarrow ab = 15 \quad \text{(2)} \] If \( ab = 15 \) and \( a^2 + 3b^2 = 52 \), where \( a \) and \( b \) are natural numbers, find \( a + b \).

Solution using Hit and Trial:

We try natural number pairs such that \( ab = 15 \).

The only valid pair is \( (a, b) = (5, 3) \).
So, the final answer is:

\[ a + b = 5 + 3 = \boxed{8} \]

Problem:

Consider the sequence: \[ t_1 = 1, \quad t_2 = -1, \quad t_n = \left( \frac{n – 3}{n – 1} \right)t_{n – 2} \quad \text{for } n \geq 3. \] Find the value of the sum: \[ \frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \cdots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}}. \]

Solution:

Let’s compute a few even terms to see the pattern:

• \( t_1 = 1 \)
• \( t_2 = -1 \)
• \( t_3 = \left( \frac{0}{2} \right) t_1 = 0 \)
• \( t_4 = \left( \frac{1}{3} \right) t_2 = \frac{-1}{3} \)
• \( t_5 = \left( \frac{2}{4} \right) t_3 = 0 \)
• \( t_6 = \left( \frac{3}{5} \right) t_4 = \left( \frac{3}{5} \cdot \frac{-1}{3} \right) = \frac{-1}{5} \)
• \( t_8 = \left( \frac{5}{7} \right) t_6 = \left( \frac{5}{7} \cdot \frac{-1}{5} \right) = \frac{-1}{7} \)

So we observe a pattern for even terms: \[ t_{2n} = \frac{-1}{2n – 1} \quad \text{for } n \geq 1 \]

Therefore: \[ \frac{1}{t_{2n}} = – (2n – 1) \] Now, compute the sum: \[ \sum_{n=1}^{1012} \frac{1}{t_{2n}} = -\sum_{n=1}^{1012} (2n – 1) = – (1 + 3 + 5 + \cdots + 2023) \] This is a sum of the first 1012 odd numbers, which equals: \[ \sum_{k=1}^{1012} (2k – 1) = 1012^2 = 1024144 \] Therefore, the final answer is: \[ \boxed{-1024144} \]