Matrices & Determinants
Determinants and Series
Grade 12
Question:
<p>If \(\alpha\), \(\beta\), \(\gamma\) are constants, then the value of</p><p>\[\sum_{r=1}^{n} S_r = \begin{vmatrix} \sum_{r=1}^{n} 2^{r-1} & \alpha & 2^{n-1} \\ \sum_{r=1}^{n} 2 \cdot 3^{r-1} & \beta & 3^{n-1} \\ \sum_{r=1}^{n} 4 \cdot 5^{r-1} & \gamma & 5^{n-1} \end{vmatrix}\]</p><p>equals</p>
Step-by-Step Solution
Key Concept: Recognize that each row follows a geometric series pattern: the first column sums evaluate to constant multiples of (aⁿ - 1)/(a - 1), while the third column contains aⁿ⁻¹. This creates a linear dependence where column 1 is a scalar multiple of column 3, making the determinant zero.
<p><strong>Step 1: Evaluate geometric series in Column 1</strong></p><p>For Row 1: Σ(r=1 to n) 2^(r-1) = (2ⁿ - 1)/(2 - 1) = 2ⁿ - 1</p><p>For Row 2: Σ(r=1 to n) 2·3^(r-1) = 2·(3ⁿ - 1)/(3 - 1) = 3ⁿ - 1</p><p>For Row 3: Σ(r=1 to n) 4·5^(r-1) = 4·(5ⁿ - 1)/(5 - 1) = 5ⁿ - 1</p><p><strong>Step 2: Recognize the pattern</strong></p><p>Matrix becomes:</p><p>|2ⁿ - 1 α 2^(n-1)|</p><p>|3ⁿ - 1 β 3^(n-1)|</p><p>|5ⁿ - 1 γ 5^(n-1)|</p><p><strong>Step 3: Identify column relationship</strong></p><p>Notice that for each row: C₁ = (aⁿ - 1) and C₃ = a^(n-1)</p><p>We can write: C₁ = a·C₃ - 1, showing linear dependence between columns</p><p>More directly: C₁ - a·C₃ is proportional across all rows, creating linear dependence</p><p><strong>Step 4: Apply determinant property</strong></p><p>Since the columns are linearly dependent (Column 1 can be expressed as a linear combination of Columns 2 and 3), the determinant equals zero.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0