The sum ∑ _ n = 1 ^ 21 (3)/((4n - 1)(4n + 3)) is equal to

The sum

\mathop{∑}_{n = 1}^{21}\dfrac{3}{\left(4n - 1\right)\left(4n + 3\right)}

is equal to

Options

A $\dfrac{7}{87}$
B $\dfrac{7}{29}$
C $\dfrac{14}{87}$
D $\dfrac{21}{29}$

Correct answer

B. $\dfrac{7}{29}$

Step-by-step solution

$\mathop{∑}_{n = 1}^{21}\dfrac{3}{\left(4n - 1\right)\left(4n + 3\right)} = \dfrac{3}{4}\mathop{∑}_{n = 1}^{21}\left(\dfrac{1}{4n - 1} - \dfrac{1}{4n + 3}\right)$
$= \dfrac{3}{4}\left[\left(\dfrac{1}{3} - \dfrac{1}{7}\right) + \left(\dfrac{1}{7} - \dfrac{1}{11}\right) + \ldots . + \left(\dfrac{1}{83} - \dfrac{1}{87}\right)\right]$
$= \dfrac{3}{4}\left[\dfrac{1}{3} - \dfrac{1}{87}\right] = \dfrac{3}{4}\dfrac{84}{261} = \dfrac{7}{29}$

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