If `D_r =|(2^(r-1),2(3^(r-1)),4(5^(r-1))),(x,y,z),(2^(n)-1,3^(n)-1,5^(n)-1)|` then `sum_(r=1)^n D_r=`

Options

A. n
B. `n^2`
C. `n^3`
D. 0

Answer

To solve the problem, we need to evaluate the determinant \( D_r \) and then find the sum from \( r = 1 \) to \( n \).

Given:
\[ D_r = \begin{vmatrix} 2^{r-1} & 2(3^{r-1}) & 4(5^{r-1}) \\ x & y & z \\ 2^n - 1 & 3^n - 1 & 5^n - 1 \end{vmatrix} \]

### Step 1: Evaluate \( D_r \)

We can expand the determinant \( D_r \) using the properties of determinants. We will use the fact that if two rows of a determinant are identical, the determinant evaluates to zero.

### Step 2: Identify the rows

The first row is:
\[ (2^{r-1}, 2(3^{r-1}), 4(5^{r-1})) \]
The third row is:
\[ (2^n - 1, 3^n - 1, 5^n - 1) \]

### Step 3: Analyze the first row

If we substitute \( r = n \) into the first row, we get:
\[ (2^{n-1}, 2(3^{n-1}), 4(5^{n-1})) \]

### Step 4: Check for identical rows

If we look at the third row:
\[ (2^n - 1, 3^n - 1, 5^n - 1) \]
We can see that as \( r \) varies from \( 1 \) to \( n \), the first row will not be identical to the third row for any \( r \).

### Step 5: Compute \( D_r \)

To compute \( D_r \), we can use the determinant expansion:
\[ D_r = 2^{r-1} \begin{vmatrix} y & z \\ 3^n - 1 & 5^n - 1 \end{vmatrix} - 2(3^{r-1}) \begin{vmatrix} x & z \\ 2^n - 1 & 5^n - 1 \end{vmatrix} + 4(5^{r-1}) \begin{vmatrix} x & y \\ 2^n - 1 & 3^n - 1 \end{vmatrix} \]

### Step 6: Sum \( D_r \) from \( r = 1 \) to \( n \)

Now we need to sum \( D_r \) from \( r = 1 \) to \( n \):
\[ \sum_{r=1}^n D_r \]

### Step 7: Analyze the result

Since the first row and the third row are not identical for any \( r \), we can conclude that the determinant \( D_r \) evaluates to zero for all \( r \):
\[ D_r = 0 \]

Thus, the sum is:
\[ \sum_{r=1}^n D_r = 0 \]

### Final Answer
The final answer is:
\[ \sum_{r=1}^n D_r = 0 \]