`{:("If",a = 1 + 2 + 4 + ..."to n terms"),(,b = 1 + 3 + 9 + ..."to n terms"),(,c =1 + 5 + 25+ ..."to n terms"):}`
then `|(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n))|=`

Options

A. `(30)^(n)`
B. `(10)^(n)`
C. `0`
D. `2^(n)+3^(n)+5^(n)`

Explanation

To solve the given problem, we need to find the determinant of the matrix formed by the expressions for \(a\), \(b\), and \(c\). Let's break this down step by step.

### Step 1: Identify the sums \(a\), \(b\), and \(c\)

1. **Sum \(a\)**:
\[ a = 1 + 2 + 4 + \ldots \text{ (to n terms)} \]
This is a geometric series with the first term \(1\) and common ratio \(2\). The sum of the first \(n\) terms of a geometric series is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Here, \(a = 1\), \(r = 2\), and \(n\) is the number of terms. Thus:
\[ a = \frac{1(2^n - 1)}{2 - 1} = 2^n - 1 \]

2. **Sum \(b\)**:
\[ b = 1 + 3 + 9 + \ldots \text{ (to n terms)} \]
This is also a geometric series with first term \(1\) and common ratio \(3\). Therefore:
\[ b = \frac{1(3^n - 1)}{3 - 1} = \frac{3^n - 1}{2} \]

3. **Sum \(c\)**:
\[ c = 1 + 5 + 25 + \ldots \text{ (to n terms)} \]
This is a geometric series with first term \(1\) and common ratio \(5\). Thus:
\[ c = \frac{1(5^n - 1)}{5 - 1} = \frac{5^n - 1}{4} \]

### Step 2: Substitute \(a\), \(b\), and \(c\) into the determinant

Now we can substitute \(a\), \(b\), and \(c\) into the determinant:
\[ \begin{vmatrix} a & 2b & 4c \\ 2 & 2 & 2 \\ 2^n & 3^n & 5^n \end{vmatrix} = \begin{vmatrix} (2^n - 1) & 2 \cdot \frac{3^n - 1}{2} & 4 \cdot \frac{5^n - 1}{4} \\ 2 & 2 & 2 \\ 2^n & 3^n & 5^n \end{vmatrix} \]

This simplifies to:
\[ \begin{vmatrix} (2^n - 1) & (3^n - 1) & (5^n - 1) \\ 2 & 2 & 2 \\ 2^n & 3^n & 5^n \end{vmatrix} \]

### Step 3: Apply row operations

We can apply row operations to simplify the determinant. Let's perform the following operations:
- Replace \(R_2\) with \(R_2 - R_1\) (subtract the first row from the second row).
- Replace \(R_1\) with \(R_1\) divided by \(2\).

This gives us:
\[ \begin{vmatrix} (2^n - 1) & (3^n - 1) & (5^n - 1) \\ 0 & 0 & 0 \\ 2^n & 3^n & 5^n \end{vmatrix} \]

### Step 4: Evaluate the determinant

Since the second row consists entirely of zeros, the determinant of this matrix is:
\[ \text{Determinant} = 0 \]

### Conclusion

Thus, the value of the determinant is:
\[ \boxed{0} \]