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Question

Mathematics Question on Determinants

If x is a positive integer, then ∣x!(x+1)!(x+2)!\[0.3em](x+1)!(x+2)!(x+3)!\[0.3em](x+2)!(x+3)!(x+4)!∣\begin{vmatrix} x! & (x+1)! & (x+2)! \\\[0.3em] (x+1)! & (x+2)! & (x+3)! \\\[0.3em] (x+2)! & (x+3)! &(x+4)! \end{vmatrix} is equal to

A

2x!(x+1)!2x ! (x + 1) !

B

2x!(x+1)!(x+2)!2x ! (x + 1)! (x + 2) !

C

2x!(x+3)!2x ! (x + 3) !

D

2(x+1)!(x+2)!(x+3)!2(x + 1) ! (x + 2) ! (x + 3) !

Answer

2x!(x+1)!(x+2)!2x ! (x + 1)! (x + 2) !

Explanation

Solution

∣x!(x+1)!(x+2)!\[0.3em](x+1)!(x+2)!(x+3)!\[0.3em](x+2)!(x+3)!(x+4)!∣\begin{vmatrix} x! & (x+1)! & (x+2)! \\\[0.3em] (x+1)! & (x+2)! & (x+3)! \\\[0.3em] (x+2)! & (x+3)! &(x+4)! \end{vmatrix} = x !(x+1)!(x+2)!x \, ! (x+1) ! (x + 2) ! ∣111 x+1x+2(x+3) (x+2)(x+1)(x+2)(x+3)(x+3)(x+4)∣\begin{vmatrix}1&1&1\\\ x+1&x+2&\left(x+3\right)\\\ \left(x+2\right)\left(x+1\right)&\left(x+2\right)\left(x+3\right)&\left(x+3\right)\left(x+4\right)\end{vmatrix} = 2x!(x+1)!(x+2)!2x ! (x + 1 ) ! ( x + 2 ) !