If `D_(r) = |(r,1,(n(n +1))/(2)),(2r -1,4,n^(2)),(2^(r -1),5,2^(n) -1)|`, then the value of `sum_(r=1)^(n) D_(r)`, is

Options

A. 0
B. 1
C. `(n(n +1) (2n+1))/(6)`
D. none of these

Explanation

To solve the problem, we need to evaluate the determinant \( D_r \) given by:

\[ D_r = \begin{vmatrix} r & 1 & \frac{n(n + 1)}{2} \\ 2r - 1 & 4 & n^2 \\ 2^{r - 1} & 5 & 2^n - 1 \end{vmatrix} \]

and then find the sum \( \sum_{r=1}^{n} D_r \).

### Step 1: Calculate the Determinant \( D_r \)

We will calculate the determinant using the formula for a \( 3 \times 3 \) determinant:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our determinant:

- \( a = r \), \( b = 1 \), \( c = \frac{n(n + 1)}{2} \)
- \( d = 2r - 1 \), \( e = 4 \), \( f = n^2 \)
- \( g = 2^{r - 1} \), \( h = 5 \), \( i = 2^n - 1 \)

Now, we can substitute these values into the determinant formula:

\[ D_r = r(4(2^n - 1) - n^2 \cdot 5) - 1((2r - 1)(2^n - 1) - n^2 \cdot 2^{r - 1}) + \frac{n(n + 1)}{2}((2r - 1) \cdot 5 - 4 \cdot 2^{r - 1}) \]

### Step 2: Simplify \( D_r \)

1. Calculate \( 4(2^n - 1) - 5n^2 \):
\[ = 4 \cdot 2^n - 4 - 5n^2 \]

2. Calculate \( (2r - 1)(2^n - 1) - n^2 \cdot 2^{r - 1} \):
\[ = (2r - 1)2^n - (2r - 1) - n^2 \cdot 2^{r - 1} \]

3. Calculate \( (2r - 1) \cdot 5 - 4 \cdot 2^{r - 1} \):
\[ = 10r - 5 - 4 \cdot 2^{r - 1} \]

Putting it all together, we can express \( D_r \) in a simplified form.

### Step 3: Evaluate \( \sum_{r=1}^{n} D_r \)

Now we need to evaluate the sum \( \sum_{r=1}^{n} D_r \).

### Step 4: Analyze the Columns of the Determinant

Notice that if we look closely at the columns of the determinant, we can see that:

- The second column is \( (1, 4, 5) \)
- The third column is \( \left( \frac{n(n + 1)}{2}, n^2, 2^n - 1 \right) \)

If we set \( n = 1 \) and check the values of the columns, we might find that the columns become linearly dependent.

### Step 5: Use Properties of Determinants

From the properties of determinants, if two columns (or rows) are the same, the determinant equals zero.

In our case, we can see that as we calculate \( D_r \) for different values of \( r \), we may find that the second and third columns become identical for certain values of \( r \).

Thus, we conclude:

\[ \sum_{r=1}^{n} D_r = 0 \]

### Final Answer

The value of \( \sum_{r=1}^{n} D_r \) is \( 0 \).

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