Let `Delta_(r)=|(2^(r-1),2(3^(r-1)),4(5^(r-1))),(alpha, beta, gamma),(2^(n)-1,3^(n)-1,5^(n)-1)|` for `r=1,2,………..n`. The `sum_(r=1)Delta_(r)` is

Options

A. independent of `alpha, beta, gamma` and n
B. independent of n only
C. depends on `alpha, beta, gamma` and n
D. independent of `alpha, beta, gamma` only

Answer

To solve the problem, we need to evaluate the sum of determinants defined as:

\[ \Delta_r = \begin{vmatrix} 2^{r-1} & 2 \cdot 3^{r-1} & 4 \cdot 5^{r-1} \\ \alpha & \beta & \gamma \\ 2^n - 1 & 3^n - 1 & 5^n - 1 \end{vmatrix} \]

for \( r = 1, 2, \ldots, n \), and then find the sum \( \sum_{r=1}^{n} \Delta_r \).

### Step 1: Evaluate the Determinant \(\Delta_r\)

The determinant can be evaluated using the properties of determinants. We can express it as:

\[ \Delta_r = 2^{r-1} \begin{vmatrix} 1 & 3^{r-1} & 2 \cdot 5^{r-1} \\ \alpha & \beta & \gamma \\ 2^n - 1 & 3^n - 1 & 5^n - 1 \end{vmatrix} \]

### Step 2: Apply the Determinant Expansion

Using the determinant expansion along the first row, we have:

\[ \Delta_r = 2^{r-1} \left( 1 \cdot \begin{vmatrix} \beta & \gamma \\ 3^n - 1 & 5^n - 1 \end{vmatrix} - 3^{r-1} \cdot \begin{vmatrix} \alpha & \gamma \\ 2^n - 1 & 5^n - 1 \end{vmatrix} + 2 \cdot 5^{r-1} \cdot \begin{vmatrix} \alpha & \beta \\ 2^n - 1 & 3^n - 1 \end{vmatrix} \right) \]

### Step 3: Simplify the Determinants

Each of the determinants can be evaluated as follows:

1. \(\begin{vmatrix} \beta & \gamma \\ 3^n - 1 & 5^n - 1 \end{vmatrix} = \beta(5^n - 1) - \gamma(3^n - 1)\)

2. \(\begin{vmatrix} \alpha & \gamma \\ 2^n - 1 & 5^n - 1 \end{vmatrix} = \alpha(5^n - 1) - \gamma(2^n - 1)\)

3. \(\begin{vmatrix} \alpha & \beta \\ 2^n - 1 & 3^n - 1 \end{vmatrix} = \alpha(3^n - 1) - \beta(2^n - 1)\)

### Step 4: Substitute Back into \(\Delta_r\)

Substituting these back into the expression for \(\Delta_r\) gives us a more complex expression, but we can note that each term involves powers of \(2\), \(3\), and \(5\) multiplied by constants \(\alpha\), \(\beta\), and \(\gamma\).

### Step 5: Sum Over \(r\)

Now we need to sum \(\Delta_r\) from \(r = 1\) to \(n\):

\[ \sum_{r=1}^{n} \Delta_r \]

This sum can be simplified using the formula for the sum of a geometric series.

### Step 6: Evaluate the Geometric Series

The sums of the form \( \sum_{r=1}^{n} 2^{r-1} \), \( \sum_{r=1}^{n} 3^{r-1} \), and \( \sum_{r=1}^{n} 5^{r-1} \) can be evaluated as:

\[ \sum_{r=1}^{n} 2^{r-1} = 2^n - 1, \quad \sum_{r=1}^{n} 3^{r-1} = 3^n - 1, \quad \sum_{r=1}^{n} 5^{r-1} = 5^n - 1 \]

### Step 7: Final Result

Thus, the final result for the sum \( \sum_{r=1}^{n} \Delta_r \) can be expressed in terms of these sums, leading us to conclude that the sum is independent of \(\alpha\), \(\beta\), and \(\gamma\).

### Conclusion

The final value of the sum \( \sum_{r=1}^{n} \Delta_r \) is a constant that does not depend on \(n\) or the parameters \(\alpha\), \(\beta\), and \(\gamma\).