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

The Correct Option is B

$\sum _{n=1}^{21}\frac{3}{(4n-1)(4n+3)}$

$=\frac{3}{4}\sum _{n=1}^{21}\frac{1}{4n-1}-\frac{1}{4n+3}$

$=\frac{3}{4}\sum _{n=1}^{21}\frac{(4n+3)-(4n-1)}{(4n-1)(4n+3)}$

$=\frac{3}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\dots .+\frac{1}{83}-\frac{1}{87}\right)$

$=\frac{3}{4}\left(\frac{1}{3}-\frac{1}{87}\right)=\frac{7}{29}$

So, The correct option is (B): $\frac7{29}$

Answer 2

\(\overset{21}{\underset{n=1}\sum} \frac3{(4n-1)(4n+3)} = \frac3{4}\overset{21}{\underset{n=1}\sum} \frac1{(4n-1)}-\frac1{(4n+3)}\)

=43​n=1∑21​(4n−1)(4n+3)(4n+3)−(4n−1)​ 

=43​n=1∑21​4n−11​−4n+31​ 
=43​(31​−71​+71​−111​+111​−….+831​−871​) 
=43​(31​−871​)=297​