The value of the sum of 10 term of the series $S_{10}=\frac{1}{2^2-1}+\frac{1}{4^2-1}+\frac{1}{6^2-1}+....$ is

$\frac{10}{21}$

10th term of the series = $\frac{1}{20^2-1}$

(Using identity a²-b² = (a+b)(a-b)) 

2²-1 = 2²-1² = (2+1)(2-1) = 3*1

Similarly use the same for other terms 

S= 1/(1*3) + 1/(3*5) + 1/(5*7)......

Since they have difference of 2 in denominator terms

Take multiply and divide by 2

S= (1/2) [2/(1*3) + 2/(3*5) + 2/(5*7)......]

S= (1/2) [(3-1)/(1*3) + (5-3)/(3*5) + (7-5)/(5*7)......]

S= (1/2) [(1/1)-(1/3) + (1/3)-(1/5) + (1/5)-(1/7)......+(1/19) - (1/21]

All the +ve and -ve terms are eliminated except 1 and 1/21

S= (1/2) (1- 1/21)

S= 0.5*20/21

The correct answer is Option (4) → $\frac{10}{21}$