If `D_r=|(r,x,(n(n+1))/2),(2r,1,n^2),(3r,-2,(n(3n-1))/2)|` then `sum_(r=1)^nD_r=0`

Answer

To solve the given problem, we need to evaluate the determinant \( D_r \) and then find the value of \( x \) such that the summation \( \sum_{r=1}^{n} D_r = 0 \).

### Step-by-Step Solution:

1. **Write the Determinant:**
\[ D_r = \begin{vmatrix} r & x & \frac{n(n+1)}{2} \\ 2r & 1 & n^2 \\ 3r & -2 & \frac{n(3n-1)}{2} \end{vmatrix} \]

2. **Factor out \( r \):**
Since \( r \) is common in the first column, we can factor it out:
\[ D_r = r \begin{vmatrix} 1 & x & \frac{n(n+1)}{2} \\ 2 & 1 & n^2 \\ 3 & -2 & \frac{n(3n-1)}{2} \end{vmatrix} \]

3. **Perform Row Operations:**
We perform row operations to simplify the determinant:
- Replace \( R_2 \) with \( R_2 - 2R_1 \)
- Replace \( R_3 \) with \( R_3 - 3R_1 \)
\[ D_r = r \begin{vmatrix} 1 & x & \frac{n(n+1)}{2} \\ 0 & 1 - 2x & n^2 - n(n+1) \\ 0 & -2 - 3x & \frac{n(3n-1)}{2} - 3 \cdot \frac{n(n+1)}{2} \end{vmatrix} \]

4. **Simplify the Second and Third Rows:**
Calculate the entries of the second and third rows:
- For the second row, the third column becomes:
\[ n^2 - \frac{n(n+1)}{2} = \frac{2n^2 - n^2 - n}{2} = \frac{n^2 - n}{2} \]
- For the third row, the third column becomes:
\[ \frac{n(3n-1)}{2} - \frac{3n(n+1)}{2} = \frac{3n^2 - n - 3n^2 - 3n}{2} = \frac{-4n}{2} = -2n \]

5. **Write the Simplified Determinant:**
Now, we have:
\[ D_r = r \begin{vmatrix} 1 & x & \frac{n(n+1)}{2} \\ 0 & 1 - 2x & \frac{n^2 - n}{2} \\ 0 & -2 - 3x & -2n \end{vmatrix} \]

6. **Calculate the Determinant:**
The determinant can be calculated using the first column:
\[ D_r = r \left( 1 \cdot \left( (1 - 2x)(-2n) - (-2 - 3x) \cdot \frac{n^2 - n}{2} \right) \right) \]
This simplifies to:
\[ D_r = r \left( -2n(1 - 2x) + (2 + 3x) \cdot \frac{n^2 - n}{2} \right) \]

7. **Set Up the Summation:**
We know that:
\[ \sum_{r=1}^{n} D_r = 0 \]
This implies:
\[ \sum_{r=1}^{n} r \left( -2n(1 - 2x) + (2 + 3x) \cdot \frac{n^2 - n}{2} \right) = 0 \]

8. **Factor Out Constants:**
Since \( n \) and \( n+1 \) are not zero, we can factor them out:
\[ n(n+1) \left( x - 4 \right) = 0 \]
This leads to:
\[ x - 4 = 0 \implies x = 4 \]

### Final Answer:
\[ \boxed{4} \]