What is the value of the following expression? $\frac{1}{(2^2 - 1)} + \frac{1}{(4^2 - 1)} + \frac{1}{(6^2 - 1)} + \cdots + \frac{1}{(20^2 - 1)}$

### Concept & Formula This series can be solved by recognizing it as a telescoping sum. We first expand the denominator using the difference of squares $a^2 - b^2 = (a - b)(a + b)$, and then use partial fractions to split the terms. $$T_k = \frac{1}{(2k)^2 - 1} = \frac{1}{(2k - 1)(2k + 1)}$$ Using partial fractions, this splits into: $$\frac{1}{(2k - 1)(2k + 1)} = \frac{1}{2} \left[ \frac{1}{2k - 1} - \frac{1}{2k + 1} \right]$$ ### Step-by-Step Solution Apply the partial fraction decomposition to each term in the series: First term ($k=1$): $\frac{1}{2^2 - 1} = \frac{1}{1 \times 3} = \frac{1}{2} \left( \frac{1}{1} - \frac{1}{3} \right)$ Second term ($k=2$): $\frac{1}{4^2 - 1} = \frac{1}{3 \times 5} = \frac{1}{2} \left( \frac{1}{3} - \frac{1}{5} \right)$ Third term ($k=3$): $\frac{1}{6^2 - 1} = \frac{1}{5 \times 7} = \frac{1}{2} \left( \frac{1}{5} - \frac{1}{7} \right)$ Last term ($k=10$): $\frac{1}{20^2 - 1} = \frac{1}{19 \times 21} = \frac{1}{2} \left( \frac{1}{19} - \frac{1}{21} \right)$ Now, add all the terms together. We can factor out the $\frac{1}{2}$ from the entire sum: Sum = $\frac{1}{2} \left[ \left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{19} - \frac{1}{21}\right) \right]$ Observe the interior terms cancelling out: $-\frac{1}{3}$ cancels $+\frac{1}{3}$, and so on. The only terms remaining are the first and the last: Sum = $\frac{1}{2} \left[ 1 - \frac{1}{21} \right]$ Sum = $\frac{1}{2} \left[ \frac{20}{21} \right] = \frac{10}{21}$ ### Exam Strategy & Shortcut When faced with a series of fractions $\frac{1}{a \times b} + \frac{1}{b \times c} + \dots + \frac{1}{y \times z}$ where the difference in the denominator factors ($b - a$) is constant (let us call this difference $D$), the sum is simply: $$\text{Sum} = \frac{1}{D} \times \left( \frac{1}{\text{First Factor}} - \frac{1}{\text{Last Factor}} \right)$$ Here, the factors are $1 \times 3, 3 \times 5, \dots, 19 \times 21$. The difference $D = 2$. First factor is $1$, last factor is $21$. $$\text{Sum} = \frac{1}{2} \left( \frac{1}{1} - \frac{1}{21} \right) = \frac{10}{21}$$ This takes less than 15 seconds to execute. ### Common Pitfall Forgetting to multiply by the factor $\frac{1}{2}$ (or $\frac{1}{D}$ in general) is the most frequent mistake. Students correctly identify the first and last terms but compute $1 - \frac{1}{21} = \frac{20}{21}$, completely missing the $\frac{1}{2}$ scaling factor. ### Final Answer **Therefore, the correct answer is 10/21.**