The sum $$\displaystyle\sum_{n=1}^{21} \dfrac{3}{(4n-1)(4n+3)}$$ is equal to

We need to find the value of $$\displaystyle\sum_{n=1}^{21} \dfrac{3}{(4n-1)(4n+3)}$$.

Decompose using partial fractions

$$\dfrac{3}{(4n-1)(4n+3)} = \dfrac{3}{4}\left(\dfrac{1}{4n-1} - \dfrac{1}{4n+3}\right)$$

This can be verified by combining the right side:

$$\dfrac{3}{4} \cdot \dfrac{(4n+3)-(4n-1)}{(4n-1)(4n+3)} = \dfrac{3}{4} \cdot \dfrac{4}{(4n-1)(4n+3)} = \dfrac{3}{(4n-1)(4n+3)} \checkmark$$

Write the telescoping sum

$$S = \dfrac{3}{4}\sum_{n=1}^{21}\left(\dfrac{1}{4n-1} - \dfrac{1}{4n+3}\right)$$

$$= \dfrac{3}{4}\left[\left(\dfrac{1}{3} - \dfrac{1}{7}\right) + \left(\dfrac{1}{7} - \dfrac{1}{11}\right) + \left(\dfrac{1}{11} - \dfrac{1}{15}\right) + \cdots + \left(\dfrac{1}{83} - \dfrac{1}{87}\right)\right]$$

Simplify the telescoping sum

Most terms cancel, leaving:

$$S = \dfrac{3}{4}\left(\dfrac{1}{3} - \dfrac{1}{87}\right)$$

Compute the result

$$S = \dfrac{3}{4} \times \dfrac{87 - 3}{3 \times 87} = \dfrac{3}{4} \times \dfrac{84}{261} = \dfrac{3}{4} \times \dfrac{28}{87} = \dfrac{84}{348} = \dfrac{7}{29}$$

The correct answer is Option B: $$\dfrac{7}{29}$$.