A) \[2(x!)(x+1)!\]
B) \[2(x!)(x+1)!(x+2)!\]
C) \[2(x!)(x+3)!\]
D) None of these
Correct Answer: B
Solution :
\[\Delta =\left| \begin{matrix} x! & (x+1)! & (x+2)\,! \\ (x+1)! & (x+2)! & (x+3)\,! \\ (x+2)! & (x+3)! & (x+4)\,! \\ \end{matrix} \right|\] = \[x\,!\,(x+1)\,!\,(x+2)\,!\,\left| \,\begin{matrix} 1 & (x+1) & (x+2)\,(x+1) \\ 1 & (x+2) & (x+3)\,(x+2) \\ 1 & (x+3) & (x+4)\,(x+3) \\ \end{matrix}\, \right|\] Applying \[{{R}_{1}}\to {{R}_{2}}-{{R}_{1}},{{R}_{2}}\to ({{R}_{3}}-{{R}_{2}})\] we get \[=x\,!(x+1)\,!(x+2)\,!\] \[\,\left| \,\begin{matrix} 0 & 1 & 2(x+2) \\ 0 & 1 & 2(x+3) \\ 1 & (x+3) & (x+4)\,(x+3) \\ \end{matrix}\, \right|\] \[=2\,x!(x+1)!(x+2)!\],(on simplification). Trick: Put \[x=1\]and then match the alternate.
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