Question

$\frac{1}{3^2-1}+\frac{1}{5^2-1}+\frac{1}{7^2-1}+......+\frac{1}{{35}^2-1}=$

• A. $\frac{4}{17}$
• B. $\frac{17}{72}$
• C. $\frac{19}{72}$
• D. $\frac{17}{36}$

Answer 1

The correct option is B $\frac{17}{72}$

given series $={\sum }_{n=1}^{17}\frac{1}{(2n+1{)}^2-1}$

$=\sum \frac{1}{4n^4+4n}=\sum \frac{1}{4n}\times \frac{1}{n+1}$

solving this by integration, we will get

$=\frac{17}{72}$