`sum_(i=1)^(n) sum_(i=1)^(n) i` is equal to

Options

A. `(n(n+1))/(2)`
B. `(n(n+1)^(2))/2`
C. `(n^(2)(n+1))/2`
D. none of these

Answer

To solve the problem, we need to evaluate the double summation \( \sum_{i=1}^{n} \sum_{j=1}^{n} j \).

### Step-by-Step Solution:

1. **Understand the Inner Summation**:
The inner summation \( \sum_{j=1}^{n} j \) represents the sum of the first \( n \) natural numbers.

2. **Calculate the Inner Summation**:
The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{j=1}^{n} j = \frac{n(n + 1)}{2} \]

3. **Substitute the Inner Summation into the Outer Summation**:
Now, we substitute the result of the inner summation back into the outer summation:
\[ \sum_{i=1}^{n} \left( \sum_{j=1}^{n} j \right) = \sum_{i=1}^{n} \left( \frac{n(n + 1)}{2} \right) \]

4. **Evaluate the Outer Summation**:
Since \( \frac{n(n + 1)}{2} \) is a constant with respect to \( i \), we can factor it out of the summation:
\[ \sum_{i=1}^{n} \left( \frac{n(n + 1)}{2} \right) = \frac{n(n + 1)}{2} \cdot \sum_{i=1}^{n} 1 \]

5. **Calculate \( \sum_{i=1}^{n} 1 \)**:
The summation \( \sum_{i=1}^{n} 1 \) simply counts the number of terms from 1 to \( n \), which is \( n \):
\[ \sum_{i=1}^{n} 1 = n \]

6. **Combine the Results**:
Now we can substitute back:
\[ \frac{n(n + 1)}{2} \cdot n = \frac{n^2(n + 1)}{2} \]

7. **Final Result**:
Thus, the value of the double summation \( \sum_{i=1}^{n} \sum_{j=1}^{n} j \) is:
\[ \frac{n^2(n + 1)}{2} \]

### Conclusion:
The final answer is:
\[ \frac{n^2(n + 1)}{2} \]