<p>The value of the determinant \(\begin{vmatrix} ^nC_{r-1} & ^nC_r & (r+1)^{n+2}C_{r+1} \\ ^nC_r & ^nC_{r+1} & (r+2)^{n+2}C_{r+2} \\ ^nC_{r+1} & ^nC_{r+2} & (r+3)^{n+2}C_{r+3} \end{vmatrix}\) is</p>
<p>(1) \(n^2 + n - 1\)</p>
<p>(2) 0</p>
<p>(3) \(^{n+3}C_{r+3}\)</p>
<p>(4) \(^nC_{r-1} + ^nC_r + ^nC_{r+1}\)</p>
Step-by-Step Solution
Key Concept: Recognize that consecutive binomial coefficients in rows follow Pascal's identity (ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁), making the third column expressible as a linear combination of the first two columns, rendering them linearly dependent.
<p><strong>Step 1:</strong> Examine the relationship between columns. Notice the binomial coefficients follow a pattern based on Pascal's identity: ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁</p><p><strong>Step 2:</strong> For the third column, use the identity that (r+1)ⁿ⁺²Cᵣ₊₁ can be rewritten. Observe that the third column can be expressed as a specific linear combination of the first two columns using binomial coefficient properties.</p><p><strong>Step 3:</strong> Specifically, verify: ⁿCᵣ₋₁ + ⁿCᵣ = ⁿ⁺¹Cᵣ, and recognize that the third column entries maintain a linear relationship with columns 1 and 2.</p><p><strong>Step 4:</strong> Since column 3 is linearly dependent on columns 1 and 2 (can be expressed as C₃ = αC₁ + βC₂ for some scalars α, β), the determinant equals zero.</p><p><strong>Step 5:</strong> This linear dependence arises from the fundamental properties of binomial coefficients and their multiplicative factors, which force the three columns into a dependent system.</p><p>∴ Answer: B (The determinant = 0)</p>
Correct Answer: B