Find the sum of `(1 - (1)/(n + 1)) + (1 - (2)/(n + 1)) + (1 - (3)/(n + 1))+......(1 - (n)/(n + 1))`

Options

A. n
B. `(1)/(2)n`
C. (n + 1)
D. `(1)/(2) (n + 1)`

Answer

To find the sum of the expression

\[ (1 - \frac{1}{n + 1}) + (1 - \frac{2}{n + 1}) + (1 - \frac{3}{n + 1}) + \ldots + (1 - \frac{n}{n + 1}), \]

we can simplify it step by step.

### Step 1: Rewrite the Expression
We can rewrite the sum as follows:

\[ \sum_{k=1}^{n} \left(1 - \frac{k}{n + 1}\right). \]

This means we are summing the terms from \(k = 1\) to \(n\).

### Step 2: Separate the Terms
Now, we can separate the sum into two parts:

\[ \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \frac{k}{n + 1}. \]

### Step 3: Calculate the First Sum
The first sum, \(\sum_{k=1}^{n} 1\), is simply \(n\) because we are adding 1 a total of \(n\) times.

\[ \sum_{k=1}^{n} 1 = n. \]

### Step 4: Calculate the Second Sum
For the second sum, we have:

\[ \sum_{k=1}^{n} \frac{k}{n + 1} = \frac{1}{n + 1} \sum_{k=1}^{n} k. \]

The sum of the first \(n\) natural numbers is given by the formula:

\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2}. \]

Thus, substituting this into our expression gives:

\[ \sum_{k=1}^{n} \frac{k}{n + 1} = \frac{1}{n + 1} \cdot \frac{n(n + 1)}{2} = \frac{n}{2}. \]

### Step 5: Combine the Results
Now we can substitute back into our separated sums:

\[ n - \frac{n}{2} = \frac{2n}{2} - \frac{n}{2} = \frac{n}{2}. \]

### Final Result
Thus, the sum of the given expression is:

\[ \frac{n}{2}. \]