Let `|Z_(r) - r| le r`, for all `r = 1,2,3….,n`. Then `|sum_(r=1)^(n)z_(r)|` is less than

Options

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

Explanation

To solve the problem, we need to analyze the given inequality and derive the required expression step by step.

### Step-by-Step Solution:

1. **Understanding the Given Inequality**:
We are given that \( |Z_r - r| \leq r \) for \( r = 1, 2, 3, \ldots, n \). This means that the complex number \( Z_r \) is within a distance \( r \) from the real number \( r \).

2. **Rearranging the Inequality**:
From the given inequality, we can express \( Z_r \) as:
\[ |Z_r| \leq |Z_r - r| + |r| \leq r + r = 2r \]
This means that the modulus of \( Z_r \) is less than or equal to \( 2r \).

3. **Summing Over r**:
Now, we want to find the modulus of the sum:
\[ \left| \sum_{r=1}^{n} Z_r \right| \]
Using the triangle inequality, we have:
\[ \left| \sum_{r=1}^{n} Z_r \right| \leq \sum_{r=1}^{n} |Z_r| \]

4. **Applying the Bound on |Z_r|**:
From our previous step, we know \( |Z_r| \leq 2r \). Therefore:
\[ \sum_{r=1}^{n} |Z_r| \leq \sum_{r=1}^{n} 2r = 2 \sum_{r=1}^{n} r \]

5. **Calculating the Sum**:
The sum \( \sum_{r=1}^{n} r \) is given by the formula:
\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]
Thus:
\[ \sum_{r=1}^{n} |Z_r| \leq 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]

6. **Final Result**:
Therefore, we conclude that:
\[ \left| \sum_{r=1}^{n} Z_r \right| \leq n(n + 1) \]

### Conclusion:
The final result is:
\[ | \sum_{r=1}^{n} Z_r | < n(n + 1) \]