Question

If n is positive integer, then $\left|\begin{array}{ccc}{n+2}_{C_n}& n+3_{C_{n+1}}& n+4_{C_{n+2}}\\ n+3_{C_{n+1}}& n+4_{C_{n+2}}& n+5_{C_{n+3}}\\ n+4_{C_{n+2}}& n+5_{C_{n+3}}& n+6_{C_{n+4}}\end{array}\right|$ is equal to

• A. 3
• B. -1
• C. -5
• D. -9

Answer 1

The correct option is B -1

$n_{C_r}=n_{C_{n-r}}$
$\therefore \left|\begin{array}{ccc}n+2_{C_2}& n+3_{C_2}& n+4_{C_2}\\ n+3_{C_2}& n+4_{C_2}& n+5_{C_2}\\ n+4_{C_2}& n+5_{C_2}& n+6_{C_2}\end{array}\right|=\left|\begin{array}{ccc}\frac{(n+2)(n+1)}{2}& \frac{(n+3)(n+2)}{2}& \frac{(n+4)(n+3)}{2}\\ \frac{(n+3)(n+2)}{2}& \frac{(n+4)(n+3)}{2}& \frac{(n+5)(n+4)}{2}\\ \frac{(n+4)(n+3)}{2}& \frac{(n+5)(n+4)}{2}& \frac{(n+6)(n+5)}{5}\end{array}\right|$
$\frac{1}{8}\left|\begin{array}{ccc}{n}^2+3n+2& n^2+5n+6& n^2+7n+12\\ n^2+5n+6& n^2+7n+12& n^2+9n+20\\ n^2+7n+12& n^2+9n+20& n^2+11n+30\end{array}\right|$
Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, then
$\frac{1}{8}\left|\begin{array}{ccc}{n}^2+3n+2& 2n+4& 4n+10\\ n^2+5n+6& 2n+6& 4n+14\\ n^2+7n+12& 2n+8& 4n+18\end{array}\right|$
Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$\frac{1}{8}\left|\begin{array}{ccc}{n}^2+3n+2& 2n+4& 4n+10\\ 2n+4& 2& 4\\ 4n+10& 4& 8\end{array}\right|$
Applying $R_3\to R_3-2R_2$
$\frac{1}{8}\left|\begin{array}{ccc}{n}^2+3n+2& 2n+4& 4n+10\\ ⋮& & \\ 2n+4& 2& 4\\ ⋮& & \\ 2& \cdots 0\cdots & 0\end{array}\right|=\frac{2}{8}\left\{\right(8n+16)-(8n+20\left)\right\}=-1$