Question

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

• A. $\frac{7}{87}$
• B. $\frac{7}{29}$
• C. $\frac{14}{87}$
• D. $\frac{21}{29}$

Answer 1

Answer: (B)

$ \displaystyle\sum_{n=1}^{21} \frac{3}{(4 n-1)(4 n+3)} $
$ =\frac{3}{4} \displaystyle\sum_{n=1}^{21} \frac{(4 n+3)-(4 n-1)}{(4 n-1)(4 n+3)} $
$=\frac{3}{4} \displaystyle\sum_{n=1}^{21} \frac{1}{4 n-1}-\frac{1}{4 n+3}$
$ =\frac{3}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\ldots .+\frac{1}{83}-\frac{1}{87}\right) $
$ =\frac{3}{4}\left(\frac{1}{3}-\frac{1}{87}\right)=\frac{7}{29}$