1. Key Concepts and Formulas

• Partial Fraction Decomposition: A technique to express a rational function as a sum of simpler rational functions. For a term of the form $\frac{P(x)}{(ax+b)(cx+d)}$, we can write it as $\frac{A}{ax+b} + \frac{B}{cx+d}$.
• Telescoping Series: A series where most of the intermediate terms cancel out when the sum is computed. A common form is $\sum_{n=1}^{N} (f(n) - f(n+1)) = f(1) - f(N+1)$.

2. Step-by-Step Solution

• Step 1: Decompose the general term using Partial Fractions. The general term of the series is $T_n = \frac{3}{(4n - 1)(4n + 3)}$. We decompose this into partial fractions: $\frac{3}{(4n - 1)(4n + 3)} = \frac{A}{4n - 1} + \frac{B}{4n + 3}$ Multiplying both sides by $(4n - 1)(4n + 3)$ gives: $3 = A(4n + 3) + B(4n - 1)$ To find $A$, set $4n - 1 = 0 \implies n = \frac{1}{4}$: $3 = A\left(4\left(\frac{1}{4}\right) + 3\right) + B\left(4\left(\frac{1}{4}\right) - 1\right)$ $3 = A(1 + 3) + B(0) \implies 3 = 4A \implies A = \frac{3}{4}$ To find $B$, set $4n + 3 = 0 \implies n = -\frac{3}{4}$: $3 = A\left(4\left(-\frac{3}{4}\right) + 3\right) + B\left(4\left(-\frac{3}{4}\right) - 1\right)$ $3 = A(0) + B(-3 - 1) \implies 3 = -4B \implies B = -\frac{3}{4}$ So, the general term is: $T_n = \frac{3/4}{4n - 1} - \frac{3/4}{4n + 3} = \frac{3}{4} \left( \frac{1}{4n - 1} - \frac{1}{4n + 3} \right)$

• Step 2: Identify the telescoping pattern. Let $f(n) = \frac{1}{4n - 1}$. Then, $f(n+1) = \frac{1}{4(n+1) - 1} = \frac{1}{4n + 4 - 1} = \frac{1}{4n + 3}$. Thus, the general term is in the form $T_n = \frac{3}{4} (f(n) - f(n+1))$. This indicates a telescoping series.

• Step 3: Calculate the sum of the series. We need to compute the sum $S = \sum_{n=1}^{21} T_n$. $S = \sum_{n=1}^{21} \frac{3}{4} \left( \frac{1}{4n - 1} - \frac{1}{4n + 3} \right)$ Factor out the constant $\frac{3}{4}$: $S = \frac{3}{4} \sum_{n=1}^{21} \left( \frac{1}{4n - 1} - \frac{1}{4n + 3} \right)$ Let's write out the terms of the sum: For $n=1$: $\frac{1}{4(1)-1} - \frac{1}{4(1)+3} = \frac{1}{3} - \frac{1}{7}$ For $n=2$: $\frac{1}{4(2)-1} - \frac{1}{4(2)+3} = \frac{1}{7} - \frac{1}{11}$ For $n=3$: $\frac{1}{4(3)-1} - \frac{1}{4(3)+3} = \frac{1}{11} - \frac{1}{15}$ ... For $n=21$: $\frac{1}{4(21)-1} - \frac{1}{4(21)+3} = \frac{1}{83} - \frac{1}{87}$

  The sum is: $\left(\frac{1}{3} - \cancel{\frac{1}{7}}\right) + \left(\cancel{\frac{1}{7}} - \cancel{\frac{1}{11}}\right) + \left(\cancel{\frac{1}{11}} - \cancel{\frac{1}{15}}\right) + \dots + \left(\cancel{\frac{1}{83}} - \frac{1}{87}\right)$ The intermediate terms cancel out, leaving: $\frac{1}{3} - \frac{1}{87}$ Combine these fractions: $\frac{29}{87} - \frac{1}{87} = \frac{28}{87}$ Now, multiply by the constant factor $\frac{3}{4}$: $S = \frac{3}{4} \times \frac{28}{87} = \frac{3 \times 28}{4 \times 87}$ Simplify the expression: $S = \frac{3 \times (4 \times 7)}{4 \times (3 \times 29)} = \frac{7}{29}$

3. Common Mistakes & Tips

• Algebraic Errors in Partial Fractions: Double-check the calculation of constants $A$ and $B$ to avoid sign errors or incorrect values.
• Forgetting the Constant Factor: Ensure that any constant factored out from the summation is multiplied back into the final result.
• Incorrectly Identifying the Telescoping Terms: Verify that the terms $f(n)$ and $f(n+1)$ are correctly identified and that the cancellation pattern is accurate.

4. Summary

The problem involves summing a series where the general term can be decomposed into a difference of two simpler fractions. By applying partial fraction decomposition, we rewrite the general term as $\frac{3}{4} \left( \frac{1}{4n - 1} - \frac{1}{4n + 3} \right)$. This form reveals a telescoping series, where intermediate terms cancel out. The sum simplifies to $\frac{3}{4}$ times the difference between the first term of the first fraction and the last term of the second fraction, which evaluates to $\frac{7}{29}$.

The final answer is $\boxed{\frac{7}{29}}$ which corresponds to option (B).