Find the value of

​$$\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+.......+\frac{1}{48×50}:$$​​

Correct option is D

Given:

​$$\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \cdots + \frac{1}{48 \times 50}$$​​

Concept Used:

​$$\frac{1}{n(n+2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right)$$​

Solution:

​$$\frac{1}{2} \left( \frac{1}{2} - \frac{1}{4} \right)+ \frac{1}{2} \left( \frac{1}{4} - \frac{1}{6} \right)+ \cdots + \left( \frac{1}{48} - \frac{1}{50} \right)$$​​

​$$=\frac{1}{2} \left( \frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{6} + \frac{1}{6} - \frac{1}{8} + \cdots + \frac{1}{48} - \frac{1}{50} \right)$$​​

​$$=\frac{1}{2} \left( \frac{1}{2} - \frac{1}{50} \right) = \frac{1}{2} \times \frac{48}{100} = \frac{24}{100} = \frac{6}{25}$$ = 0.24