The value of `(1)/(1xx2)+(1)/(2xx3)+(1)/(3xx4)+..... +(1)/(99xx100)` is

Options

A. less than `(99)/(100)`
B. equal to `(99)/(100)`
C. equal to `(99)/(100)`
D. greater than `(100)/(99)`

Answer

To solve the problem, we need to evaluate the sum:

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

### Step 1: Rewrite each term
We can rewrite each term in the sum using the formula:

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

This means we can express each term as follows:

\[ \frac{1}{1 \times 2} = 1 - \frac{1}{2} \]
\[ \frac{1}{2 \times 3} = \frac{1}{2} - \frac{1}{3} \]
\[ \frac{1}{3 \times 4} = \frac{1}{3} - \frac{1}{4} \]
\[ \vdots \]
\[ \frac{1}{99 \times 100} = \frac{1}{99} - \frac{1}{100} \]

### Step 2: Write the entire sum
Now, we can write the entire sum \( S \) as:

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

### Step 3: Observe the cancellation
Notice that in this series, all intermediate terms cancel out:

- The \( -\frac{1}{2} \) from the first term cancels with the \( +\frac{1}{2} \) from the second term.
- The \( -\frac{1}{3} \) from the second term cancels with the \( +\frac{1}{3} \) from the third term.
- This pattern continues all the way to \( -\frac{1}{99} \) from the second last term cancelling with \( +\frac{1}{99} \) from the last term.

### Step 4: Write the remaining terms
After cancellation, we are left with:

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

### Step 5: Simplify the expression
Now, simplify the expression:

\[ S = 1 - \frac{1}{100} = \frac{100}{100} - \frac{1}{100} = \frac{99}{100} \]

### Final Answer
Thus, the value of the sum is:

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