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For n ∈ N, evaluate the determinant D = | n! (n+1)! (n+2)! | | (n+1)! (n+2)! (n+3)! | | (n+2)! (n+3)! (n+4)! |.

1. (n!)² (2n³ − 8n²)
2. (2n!)³ (3n² + 4n − 5)
3. (n!)³ (2n³ + 8n² + 10n + 4)
4. None of these

Correct answer: (n!)³ (2n³ + 8n² + 10n + 4)

Solution

Taking out n!, (n+1)!, (n+2)! from the rows and reducing the resulting determinant gives D = (n!)^3 (2n^3 + 8n^2 + 10n + 4) (verified for n=1,2,3).

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