Let `D_(r) = |(a,2^(r),2^(16) -1),(b,3(4^(r)),2(4^(16) -1)),(c,7(8^(r)),4(8^(16) -1))|`, then the value of `underset(k =1)overset(16)Sigma D_(k)`, is (a) 0 (b) `a+b+c` (c)`ab+bc+ca` (d) 1

Options

A. 0
B. `a+b+c`
C. `ab+bc+ca`
D. `1`

Explanation

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

\[ D_r = \begin{vmatrix} a & 2^r & 2^{16} - 1 \\ b & 3 \cdot 4^r & 2 \cdot 4^{16} - 1 \\ c & 7 \cdot 8^r & 4 \cdot 8^{16} - 1 \end{vmatrix} \]

and then find the sum \( \sum_{k=1}^{16} D_k \).

### Step 1: Write the determinant for \( D_k \)

The determinant \( D_k \) can be expressed as:

\[ D_k = \begin{vmatrix} a & 2^k & 2^{16} - 1 \\ b & 3 \cdot 4^k & 2 \cdot 4^{16} - 1 \\ c & 7 \cdot 8^k & 4 \cdot 8^{16} - 1 \end{vmatrix} \]

### Step 2: Expand the determinant

Using the properties of determinants, we can expand \( D_k \):

\[ D_k = a \begin{vmatrix} 3 \cdot 4^k & 2 \cdot 4^{16} - 1 \\ 7 \cdot 8^k & 4 \cdot 8^{16} - 1 \end{vmatrix} - 2^k \begin{vmatrix} b & 2 \cdot 4^{16} - 1 \\ c & 4 \cdot 8^{16} - 1 \end{vmatrix} + (2^{16} - 1) \begin{vmatrix} b & 3 \cdot 4^k \\ c & 7 \cdot 8^k \end{vmatrix} \]

### Step 3: Analyze the structure of the determinant

Notice that the second column of the determinant \( D_k \) contains terms that are functions of \( k \). When we sum \( D_k \) from \( k = 1 \) to \( k = 16 \), the resulting sums will involve sums of powers of 2, 4, and 8.

### Step 4: Calculate the sums

1. **Sum of \( 2^k \)** from \( k=1 \) to \( 16 \):
\[ \sum_{k=1}^{16} 2^k = 2(2^{16} - 1) = 2^{17} - 2 \]

2. **Sum of \( 3 \cdot 4^k \)** from \( k=1 \) to \( 16 \):
\[ \sum_{k=1}^{16} 3 \cdot 4^k = 3(4(4^{16} - 1)/3) = 4^{17} - 4 \]

3. **Sum of \( 7 \cdot 8^k \)** from \( k=1 \) to \( 16 \):
\[ \sum_{k=1}^{16} 7 \cdot 8^k = 7(8(8^{16} - 1)/7) = 8^{17} - 8 \]

### Step 5: Substitute back into the determinant

After calculating the sums, we substitute back into the determinant structure. However, we notice that the second column becomes linearly dependent on the third column due to the nature of the sums.

### Step 6: Conclusion

Since two columns of the determinant become equal (or linearly dependent), the determinant \( D_k \) evaluates to zero for each \( k \).

Thus, the sum \( \sum_{k=1}^{16} D_k = 0 \).

### Final Answer

The value of \( \sum_{k=1}^{16} D_k \) is \( 0 \).