Question
Let \[f\left( x \right) = \left| {\begin{array}{*{20}{c}}
n&{n + 1}&{n + 2}\\
{{\,^n}{P_n}}&{^{n + 1}{P_{n + 1}}}&{^{n + 2}{P_{n + 2}}}\\
{{\,^n}{C_n}}&{^{n + 1}{C_{n + 1}}}&{^{n + 2}{C_{n + 2}}}
\end{array}} \right|,\] where the symbols have their usual meanings. The $$f\left( x \right)$$ is divisible by
A.
$${{n^2} + n + 1}$$
B.
$$\left( {n + 1} \right)!$$
C.
$$\left( {2n + 1} \right)!$$
D.
None of the above
Answer :
$${{n^2} + n + 1}$$
Solution :
\[\begin{gathered}
\because f\left( x \right) = \left| {\begin{array}{*{20}{c}}
n&{n + 1}&{n + 2} \\
{{\,^n}{P_n}}&{^{n + 1}{P_{n + 1}}}&{^{n + 2}{P_{n + 2}}} \\
{{\,^n}{C_n}}&{^{n + 1}{C_{n + 1}}}&{^{n + 2}{C_{n + 2}}}
\end{array}} \right| \hfill \\
= \,\left| {\begin{array}{*{20}{c}}
n&{n + 1}&{n + 2} \\
{n!}&{\left( {n + 1} \right)!}&{\left( {n + 2} \right)!} \\
1&1&1
\end{array}} \right|\,\left( {\because {\,^n}{P_n} = n!{,^n}{C_n} = 1} \right) \hfill \\
\end{gathered} \]
Applying, $${C_2} \to {C_2} - {C_1}\,\,{\text{and}}\,\,{C_3} \to {C_3} - {C_1}$$
Then, \[f\left( x \right) = \left| {\begin{array}{*{20}{c}}
n&1&2\\
{n!}&{n.n!}&{\left( {{n^2} + 3n + 1} \right)n!}\\
1&0&0
\end{array}} \right|\]
\[ = \,\left| {\begin{array}{*{20}{c}}
1&2\\
{n.n!}&{\left( {{n^2} + 3n + 1} \right)n!}
\end{array}} \right| = n!\left( {{n^2} + n + 1} \right)\]