Sequence of numbers $a_{1},a_{2},a_{3},\ldots,a_{n}$
where $a_{n}=\frac{1}{n}-\frac{1}{n+2}$ for each integer $n>0$
Now we find the $1^{st}$ term, $2^{nd}$ term up to $50^{th}$ term
$a_{1}=\frac{1}{1}-\frac{1}{1+2}=1-\frac{1}{3}$
$a_{2}=\frac{1}{2}-\frac{1}{2+2}=\frac{1}{2}-\frac{1}{4}$
$a_{3}=\frac{1}{3}-\frac{1}{3+2}=\frac{1}{3}-\frac{1}{5}$
$a_{3}=\frac{1}{4}-\frac{1}{4+2}=\frac{1}{4}-\frac{1}{6}$
$a_{50}=\frac{1}{50}-\frac{1}{50+2}=\frac{1}{50}-\frac{1}{52}$
Sum of the first $50$ terms
$S=a_{1}+a_{2}+a_{3}+a_{4}+\ldots+a_{50}$
$S=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\ldots+\frac{1}{50}-\frac{1}{52}$
We can write like this
$S=\left (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+…..+\frac{1}{50} \right )-\left ( \frac{1}{3}+\frac{1}{4}+\frac{1}{5} +\frac{1}{6}+\ldots+\frac{1}{52}\right )$
$\quad=\left (1+\frac{1}{2}\right )+ \left \{\left (\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{50} \right ) -\left ( \frac{1}{3}+\frac{1}{4}+\frac{1}{5} +\frac{1}{6}+\ldots+\frac{1}{50}\right ) \right \}-\frac{1}{51}-\frac{1}{52}$
$\quad=\left ( 1+\frac{1}{2}\right )-\left (\frac{1}{51}+\frac{1}{52} \right )$
So, correct options is (C).