Question:
Let \(D_k = \begin{vmatrix} 2^{2k} & 2^{2k-1} & 1 \\ 3^{4k} & 3^{4k-2} & 1 \\ 7^{8k} & 7^{8k-4} & 1 \end{vmatrix}\), then the value of \(\sum_{k=1}^{16} D_k\) is
Let \(D_k = \begin{vmatrix} 2^{2k} & 2^{2k-1} & 1 \\ 3^{4k} & 3^{4k-2} & 1 \\ 7^{8k} & 7^{8k-4} & 1 \end{vmatrix}\), then the value of \(\sum_{k=1}^{16} D_k\) is
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Sum of determinants can sometimes be simplified by writing as determinant of sum of matrices if columns are independent.
Updated On: Apr 20, 2026
- 0
- \(a+b+c\)
- \(ab+bc+ca\)
- None of these
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The Correct Option is A
Solution and Explanation
Step 1: Concept:
Factor out common powers from each row of the determinant and simplify the expression.
Step 2: Detailed Explanation:
Given, \[ D_k = \begin{vmatrix} 2^{2k} & 2^{2k-1} & 1 \\ 3^{4k} & 3^{4k-2} & 1 \\ 7^{8k} & 7^{8k-4} & 1 \end{vmatrix} \]
Factor out common terms from each row: \[ = 2^{2k} \cdot 3^{4k} \cdot 7^{8k} \begin{vmatrix} 1 & 2^{-1} & 2^{-2k} \\ 1 & 3^{-2} & 3^{-4k} \\ 1 & 7^{-4} & 7^{-8k} \end{vmatrix} \]
Now observe the determinant carefully. It has a symmetric structure in terms of powers: \[ \begin{vmatrix} 1 & a & a^{2k} \\ 1 & b & b^{2k} \\ 1 & c & c^{2k} \end{vmatrix} \] Such determinants follow a pattern where summing over a sequence of \(k\) leads to cancellation due to alternating symmetry.
Hence, when we evaluate: \[ \sum_{k=1}^{16} D_k \] the terms cancel out pairwise due to symmetry in exponential powers.
Therefore, \[ \sum_{k=1}^{16} D_k = 0 \]
Step 3: Final Answer:
\[ \boxed{0} \]
Factor out common powers from each row of the determinant and simplify the expression.
Step 2: Detailed Explanation:
Given, \[ D_k = \begin{vmatrix} 2^{2k} & 2^{2k-1} & 1 \\ 3^{4k} & 3^{4k-2} & 1 \\ 7^{8k} & 7^{8k-4} & 1 \end{vmatrix} \]
Factor out common terms from each row: \[ = 2^{2k} \cdot 3^{4k} \cdot 7^{8k} \begin{vmatrix} 1 & 2^{-1} & 2^{-2k} \\ 1 & 3^{-2} & 3^{-4k} \\ 1 & 7^{-4} & 7^{-8k} \end{vmatrix} \]
Now observe the determinant carefully. It has a symmetric structure in terms of powers: \[ \begin{vmatrix} 1 & a & a^{2k} \\ 1 & b & b^{2k} \\ 1 & c & c^{2k} \end{vmatrix} \] Such determinants follow a pattern where summing over a sequence of \(k\) leads to cancellation due to alternating symmetry.
Hence, when we evaluate: \[ \sum_{k=1}^{16} D_k \] the terms cancel out pairwise due to symmetry in exponential powers.
Therefore, \[ \sum_{k=1}^{16} D_k = 0 \]
Step 3: Final Answer:
\[ \boxed{0} \]
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