The series can be written as:
$\displaystyle \begin{align}S &= \sum_{i=1}^{100} \frac{1}{i \cdot (i+1)}\\[2em] &= \sum_{i=1}^{100} \left ( \frac{1}{i} - \frac{1}{i+1}\right ) \end{align}$
We have written it in the form of a telescoping series.
Now, by expanding and cancelling terms we get:
$$\require{cancel} \begin{align}S &= \left ( \frac1 1 - \frac1 2 \right ) + \left ( \frac1 2 - \frac1 3 \right ) + \cdots + \left ( \frac1{99} - \frac1{100} \right ) + \left ( \frac1{100} - \frac1{101} \right )\\[3em] &= \frac1 1 + \cancelto{0}{\left ( \frac1 2 - \frac1 2\right )} + \cancelto{0}{\left ( \frac1 3 - \frac1 3\right )}+ \cdots + \cancelto{0}{\left ( \frac1{100} - \frac1{100}\right )} - \frac1{101}\\[3em]S &= \frac1 1 - \frac1{101}\\[1.5em] &= \frac{100}{101}\end{align}$$
Hence, option d) None of these is the correct answer.