Question

If $f\left(n\right)=\left|\begin{array}{ccc}n& n+1& n+2\\ {{}^{n}P}_n& {{}^{n+1}P}_{n+1}& {{}^{n+2}P}_{n+2}\\ {{}^{n}C}_n& {{}^{n+1}C}_{n+1}& {{}^{n+2}C}_{n+2}\end{array}\right|$, where the symbols have their usual meanings. Then $f\left(n\right)$ is divisible by

• A. $n^2+n+1$
• B. $(n+1)!$
• C. $n!$
• D. none of these

Answer 1

Answer: (A), (C)

$n!$
$n^2+n+1$

$f\left(n\right)=\left|\begin{array}{ccc}n& n+1& n+2\\ {}^n{P}_n& {}^{n+1}{P}_{n+1}& {}^{n+2}{P}_{n+2}\\ {}^n{C}_n& {}^{n+1}{C}_{n+1}& {}^{n+2}{C}_{n+2}\end{array}\right|$

$=\left|\begin{array}{ccc}n& (n+1)& (n+2)\\ n!& (n+1)!& (n+2)!\\ 1& 1& 1\end{array}\right|\frac{\Lambda !}{\Lambda !\times 0!}=1$

$n((n+1)!-(n+2)!)-(n+1)(n!-(n+2)!)+(n+2)(n!-(n+1)!)$

$n(n+1)!(1-n-2)$

$-n(n+1)!(n+1)-n!(n+1)(1-(n+2)(n+1))+(n+2)(n!)(1-n+1)$

$\Rightarrow n![-{(n+1)}^{2}n+(n+1)(n^2+3n+1)+(n+2)(-n)]$

$=n![-n(n^2+2n+1)+n^3+3n^2+n+n^2+3n+1-n^2-2n]$

$=n![-n^3-2n^2-n+n^3+3n^2+n+n^2+3n+1-n^2-2n]$

$=n!(n^2+n+1)$

Option A and C