​$$750 \frac13+750 \frac1{15}+750 \frac1{35}+…750 \frac1{99}=?$$​​

Correct option is A

Given:
The series is $$750\frac{1}{3} + 750\frac{1}{15} + 750\frac{1}{35} + ... + 750\frac{1}{99}$$​​
Formula Used:
Decomposition of mixed fractions: $$a\frac{b}{c} = a + \frac{b}{c}$$​​
Telescoping series sum for terms of type $$\frac{1}{n(n+2)} = \frac{1}{2}(\frac{1}{n} - \frac{1}{n+2})$$​​
Solution:
​$$750 \left(1 +\frac{1}{3} +1+ \frac{1}{15} +1+ \frac{1}{35} +1+\frac{1}{63} +1+ \frac{1}{99}\right)$$​​

​$$= 750 \left(5+\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \ldots + \frac{1}{9 \times 11}\right)$$​

​$$= 750 \left[5+\frac{1}{2}\left(1 - \frac{1}{3}\right) + \frac{1}{2}\left(\frac{1}{3} - \frac{1}{5}\right) + \frac{1}{2}\left(\frac{1}{5} - \frac{1}{7}\right) + \ldots + \frac{1}{2}\left(\frac{1}{9} - \frac{1}{11}\right)\right]$$​

​$$= 750 \times \left[5+ \frac{1}{2}\left(1 - \frac{1}{11}\right) \right]$$​​

​$$= 750 \times\left(5+ \frac{1}{2} \times \frac{10}{11}\right)$$​​

​$$= 750 \left(5+ \frac{5}{11}\right)$$ 

​$$=3750\frac{5}{11}$$​​
Final Answer
So the correct answer is (a)