Question:

Step 1: Identify the pattern in the denominators. We observe the denominators: $9, 18, 30, 45, 63, \ldots$ Let us find the differences between consecutive terms: $$18-9=9, \quad 30-18=12, \quad 45-30=15, \quad 63-45=18$$ The differences form an arithmetic progression: $9, 12, 15, 18, \ldots$ with first term $9$ and common difference $3$. Since the first differences form an AP, the denominators follow a quadratic pattern. Step 2: Find the general term for the denominator. Let $T_n = an^2 + bn + c$ be the $n$-th denominator. Using the first three terms: - $T_1 = 9$: $a + b + c = 9$ - $T_2 = 18$: $4a + 2b + c = 18$ - $T_3 = 30$: $9a + 3b + c = 30$ Subtracting the first equation from the second: $$3a + b = 9$$ Subtracting the second equation from the third: $$5a + b = 12$$ Solving these two equations: $$2a = 3 \implies a = \frac{3}{2}$$ $$b = 9 - 3 \cdot \frac{3}{2} = 9 - \frac{9}{2} = \frac{9}{2}$$ $$c = 9 - \frac{3}{2} - \frac{9}{2} = 3$$ Therefore: $$T_n = \frac{3}{2}n^2 + \frac{9}{2}n + 3 = \frac{3(n^2 + 3n + 2)}{2} = \frac{3(n+1)(n+2)}{2}$$ Step 3: Verify the formula. Let us check: - $n=1$: $T_1 = \frac{3 \cdot 2 \cdot 3}{2} = 9$ ✓ - $n=2$: $T_2 = \frac{3 \cdot 3 \cdot 4}{2} = 18$ ✓ - $n=3$: $T_3 = \frac{3 \cdot 4 \cdot 5}{2} = 30$ ✓ Step 4: Express the general term using partial fractions. The general term of the series is: $$\frac{1}{T_n} = \frac{2}{3(n+1)(n+2)}$$ Using partial fraction decomposition: $$\frac{2}{3(n+1)(n+2)} = \frac{2}{3}\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$$ Step 5: Evaluate the infinite sum using telescoping. $$\sum_{n=1}^{\infty} \frac{1}{T_n} = \frac{2}{3} \sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+2}\right)$$ Writing out the partial sums: $$= \frac{2}{3}\left[\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \cdots\right]$$ This is a telescoping series where consecutive terms cancel: $$= \frac{2}{3}\left(\frac{1}{2}\right) = \frac{1}{3}$$ **Final Answer:** The sum of the infinite series is $\boxed{\frac{1}{3}}$, which corresponds to **Option 1**. <div class="key-concept"><strong>Key Concept:</strong> Telescoping series using partial fractions</div> <div class="trap-box"><strong>Trap:</strong> Identifying the general term of the denominator as a quadratic in n is the key step.</div>