Let `|Z_(r) - r| le r, Aar = 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)`

Answer

To solve the problem, we start with the given condition:

\[ |Z_r - r| \leq r \quad \text{for } r = 1, 2, 3, \ldots, n \]

### Step 1: Rewrite the inequality
From the inequality \( |Z_r - r| \leq r \), we can deduce that:

\[ -Z_r + r \leq |Z_r - r| \leq Z_r - r \]

This implies:

\[ -r \leq Z_r - r \leq r \]

### Step 2: Rearranging the inequality
We can rearrange the above inequalities to find bounds for \( Z_r \):

\[ 0 \leq Z_r \leq 2r \]

### Step 3: Summing over r
Now we need to find the sum:

\[ \left| \sum_{r=1}^{n} Z_r \right| \]

Using the bounds we found for \( Z_r \):

\[ \left| \sum_{r=1}^{n} Z_r \right| \leq \sum_{r=1}^{n} |Z_r| \leq \sum_{r=1}^{n} 2r \]

### Step 4: Simplifying the sum
The sum \( \sum_{r=1}^{n} r \) can be computed using the formula for the sum of the first \( n \) natural numbers:

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

Thus,

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

### Step 5: Final result
Therefore, we have:

\[ \left| \sum_{r=1}^{n} Z_r \right| \leq n(n + 1) \]

### Conclusion
The final result is:

\[ | \sum_{r=1}^{n} Z_r | < n(n + 1) \]

### Answer
The answer is \( n(n + 1) \).

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