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)

$\frac{7}{29}$

Answer 2

n=1∑21​(4n−1)(4n+3)3​

=43​n=1∑21​4n−11​−4n+31​

=43​n=1∑21​(4n−1)(4n+3)(4n+3)−(4n−1)​

=43​(31​−71​+71​−111​+111​−….+831​−871​)

=43​(31​−871​)=297​

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