The value of $\frac{3}{1^2 \cdot 2^2} + \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \frac{9}{4^2 \cdot 5^2} + \frac{11}{5^2 \cdot 6^2} + \frac{13}{6^2 \cdot 7^2} + \frac{15}{7^2 \cdot 8^2} + \frac{17}{8^2 \cdot 9^2} + \frac{19}{9^2 \cdot 10^2}$ is

### Concept & Formula This complex-looking sequence is a cleverly disguised telescoping series. Notice the relationship between the numerator and the factors in the denominator: the numerator is exactly the difference between the two squared factors in the denominator. Mathematically, the general term can be written and split as follows: $$T_n = \frac{(n+1)^2 - n^2}{n^2 \cdot (n+1)^2} = \frac{(n+1)^2}{n^2 \cdot (n+1)^2} - \frac{n^2}{n^2 \cdot (n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$$ ### Step-by-Step Solution Apply this splitting logic to every term in the given series: First term: $\frac{3}{1^2 \cdot 2^2} = \frac{4 - 1}{1^2 \cdot 2^2} = \frac{1}{1^2} - \frac{1}{2^2}$ Second term: $\frac{5}{2^2 \cdot 3^2} = \frac{9 - 4}{2^2 \cdot 3^2} = \frac{1}{2^2} - \frac{1}{3^2}$ Third term: $\frac{7}{3^2 \cdot 4^2} = \frac{16 - 9}{3^2 \cdot 4^2} = \frac{1}{3^2} - \frac{1}{4^2}$ $$\vdots$$ Last term: $\frac{19}{9^2 \cdot 10^2} = \frac{100 - 81}{9^2 \cdot 10^2} = \frac{1}{9^2} - \frac{1}{10^2}$ Now, write out the sum of all these rewritten terms: $$\text{Sum} = \left(\frac{1}{1^2} - \frac{1}{2^2}\right) + \left(\frac{1}{2^2} - \frac{1}{3^2}\right) + \dots + \left(\frac{1}{9^2} - \frac{1}{10^2}\right)$$ Every adjacent fractional pair cancels out ($-\frac{1}{2^2}$ cancels $+\frac{1}{2^2}$, and so on). The only remaining components are the first positive fraction and the last negative fraction: $$\text{Sum} = \frac{1}{1^2} - \frac{1}{10^2}$$ $$\text{Sum} = 1 - \frac{1}{100}$$ $$= \frac{99}{100}$$ ### Exam Strategy & Shortcut Whenever you see a series where the numerator equals the difference of the denominator's factors (e.g., $Numerator = b - a$ for a denominator $a \cdot b$), the entire series evaluates to: $$\text{Sum} = \frac{1}{\text{First Factor}} - \frac{1}{\text{Last Factor}}$$ Here, the first factor is $1^2 = 1$ and the last factor is $10^2 = 100$. So, $\text{Sum} = \frac{1}{1} - \frac{1}{100} = \frac{99}{100}$. You can solve this visually in under 5 seconds. ### Common Pitfall Students frequently overlook the relationship between the numerator and denominator, assuming the problem requires massive brute-force fraction addition, leading them to either skip the question or waste minutes calculating huge LCMs. ### Final Answer **Therefore, the correct answer is 99/100.**