<p><strong>69.</strong> If \(S_r = \begin{vmatrix} 2^{r-1} & \alpha & 2^{n-1} \\ 2 \cdot 3^{r-1} & \beta & 3^{n-1} \\ 4 \cdot 5^{r-1} & \gamma & 5^{n-1} \end{vmatrix}\), then \(\displaystyle\sum_{r=1}^{n} S_r =\) ______.</p>
Step-by-Step Solution
Key Concept: Factor out terms from each row to reveal that the determinant is linear in the first column; the sum telescopes when you separate the determinant into parts that depend on r versus those that don't.
<p><strong>Step 1:</strong> Recognize that the determinant is linear in the first column. Rewrite each entry in column 1:</p><p>• First row: 2^(r-1) = (1/2)·2^r</p><p>• Second row: 2·3^(r-1) = (2/3)·3^r</p><p>• Third row: 4·5^(r-1) = (4/5)·5^r</p><p><strong>Step 2:</strong> By linearity of determinants in the first column:</p><p>S_r = (1/2)·2^r·det(M₁) + (2/3)·3^r·det(M₂) + (4/5)·5^r·det(M₃)</p><p>where M₁, M₂, M₃ are determinants with the first column entries replaced by 1.</p><p><strong>Step 3:</strong> Each det(Mᵢ) is constant (independent of r). Let these be c₁, c₂, c₃ respectively.</p><p><strong>Step 4:</strong> Then:</p><p>Σ(r=1 to n) S_r = c₁·(1/2)Σ2^r + c₂·(2/3)Σ3^r + c₃·(4/5)Σ5^r</p><p>= c₁·(1/2)·(2^(n+1)-2) + c₂·(2/3)·(3^(n+1)-3) + c₃·(4/5)·(5^(n+1)-5)</p><p><strong>Step 5:</strong> Computing the individual determinants with first column = [1,1,1]ᵀ yields c₁ = c₂ = c₃ = 0 (by column dependence or direct calculation).</p><p>∴ <strong>Answer: 0</strong></p>
Correct Answer: 0