The sum of the series :
`S = 1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + …. + 1/(99.100)` is:

Options

A. `a) 98/99`
B. b)`99/100`
C. `c)S gt 1`
D. d)none of these

Answer

To find the sum of the series

\[ S = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{99 \cdot 100} \]

we can rewrite each term in a more manageable form.

### Step 1: Rewrite each term
Notice that each term can be expressed as:

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

This means we can rewrite the series as:

\[ S = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{99} - \frac{1}{100} \right) \]

### Step 2: Expand the series
When we expand this series, we can see that it forms a telescoping series:

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

### Step 3: Simplify the expression
Now, simplifying the above expression gives:

\[ S = 1 - 0.01 = 0.99 \]

### Step 4: Final answer
Thus, the sum of the series is:

\[ S = \frac{99}{100} \]