What is the sum of the following series ? `1/(1 xx 2) + 1/(2 xx 3) + 1/(3 xx 4) + ………+ 1/(100 xx 101)`

Options

A. `1/100`
B. `100/101`
C. `1/101`
D. `101/100`

Answer

To find the sum of the series

\[ S = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{100 \times 101} \]

we can rewrite each term in the series. Notice that:

\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]

This means we can express the series as:

\[ S = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{100} - \frac{1}{101} \right) \]

Now, if we expand this series, we can see that it is a telescoping series:

\[ S = 1 - \frac{1}{101} \]

The intermediate terms will cancel out, leaving us with:

\[ S = 1 - \frac{1}{101} = \frac{101}{101} - \frac{1}{101} = \frac{100}{101} \]

Thus, the sum of the series is:

\[ \boxed{\frac{100}{101}} \]