<p>If \(A_r = \begin{vmatrix} 2^{r-1} & 2 \times 3^{r-1} & 4 \times 5^{r-1} \\ \alpha & \beta & \gamma \\ 2^n - 1 & 3^n - 1 & 5^n - 1 \end{vmatrix}\), then find the value of \(\displaystyle\sum_{r=1}^{n} A_r\).</p>
Step-by-Step Solution
Key Concept: Recognize that the sum of determinants equals the determinant of the sum of matrices (when the structure is preserved). The first and third rows are fixed across all r, while the second row is constant—the sum telescopes because the first row contains geometric series terms.
<p><strong>Step 1:</strong> Use linearity of determinant in rows. Since rows 2 and 3 are constant across all A_r, we have:</p><p>$$\sum_{r=1}^{n} A_r = \begin{vmatrix} \sum_{r=1}^{n}2^{r-1} & \sum_{r=1}^{n}2 \times 3^{r-1} & \sum_{r=1}^{n}4 \times 5^{r-1} \\ \alpha & \beta & \gamma \\ 2^n - 1 & 3^n - 1 & 5^n - 1 \end{vmatrix}$$</p><p><strong>Step 2:</strong> Evaluate geometric series in row 1:</p><p>$$\sum_{r=1}^{n}2^{r-1} = \frac{2^n-1}{2-1} = 2^n-1$$</p><p>$$\sum_{r=1}^{n}2 \times 3^{r-1} = 2 \cdot \frac{3^n-1}{3-1} = 3^n-1$$</p><p>$$\sum_{r=1}^{n}4 \times 5^{r-1} = 4 \cdot \frac{5^n-1}{5-1} = 5^n-1$$</p><p><strong>Step 3:</strong> The matrix becomes:</p><p>$$\begin{vmatrix} 2^n-1 & 3^n-1 & 5^n-1 \\ \alpha & \beta & \gamma \\ 2^n-1 & 3^n-1 & 5^n-1 \end{vmatrix}$$</p><p><strong>Step 4:</strong> Row 1 and Row 3 are identical, so the determinant equals 0.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0