A) \[{{n}^{2}}+n+1\]
B) \[(n+1)!\]
C) \[(2n+1)!\]
D) None of the above
Correct Answer: A
Solution :
[a] \[\because \,\,\,f(x)=\left| \begin{matrix} n & n+1 & n+2 \\ ^{n}{{P}_{n}} & ^{n+1}{{P}_{n+1}} & ^{n+2}{{P}_{n+2}} \\ ^{n}{{C}_{n}} & ^{n+1}{{C}_{n+1}} & ^{n+2}{{C}_{n+2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} n & n+1 & n+2 \\ n! & (n+1)! & (n+2)! \\ 1 & 1 & 1 \\ \end{matrix} \right|\] \[(\because \,\,{{\,}^{n}}{{P}_{n}}=n!,\,{{\,}^{n}}{{C}_{n}}=1)\] Applying \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] Then, \[f(x)=\left| \begin{matrix} n & 1 & 2 \\ n! & n.n! & ({{n}^{2}}+3n+1)n! \\ 1 & 0 & 0 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & 2 \\ n.n! & ({{n}^{2}}+3n+1)n! \\ \end{matrix} \right|=n!({{n}^{2}}+n+1)\]
You need to login to perform this action.
You will be redirected in
3 sec
