A) \[2.x!(x+1)!(x+2)!\]
B) \[2.(x+1)!(x+2)!(x+3)!\]
C) \[2x!(x+3)!\]
D) \[2x!(x+1)!\]
Correct Answer: A
Solution :
Given, \[=\left| \begin{matrix} x! & (x+1)! & (x+2)! \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! \\ \end{matrix} \right|\] \[=\left| \begin{matrix} x! & (x+1)! & (x+2)(x+1)x! \\ (x+1)! & (x+2)(x+1)! & (x+3)(x+2)(x+1)! \\ (x+2)! & (x+3)(x+2)! & (x+4)(x+3)(x+2)! \\ \end{matrix} \right|\] \[=x!(x+1)!(x+2)!\left| \begin{matrix} 1 & x+1 & (x+2)(x+1) \\ 1 & x+2 & (x+3)(x+2) \\ 1 & x+3 & (x+3)(x+4) \\ \end{matrix} \right|\] \[=x!(x+1)!(x+2)!\left| \begin{matrix} 1 & x+1 & (x+2)(x+3) \\ 0 & 1 & 2(x+2) \\ 0 & 1 & 2(x+3) \\ \end{matrix} \right|\] \[\left[ \begin{align} & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-{{R}_{2}} \\ \end{align} \right]\] \[=x!(x+1)!(x+2)![1\{2(x+3)-2(x+2)\}]\] \[=x!(x+1)!(x+2)![2a+6-2x-4]\] \[=2.x!(x+1)!(x+2)!\]
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