Question

The value of $\left|\begin{array}{ccc}1& 1& 1\\ {}^n{C}_1& {}^{n+2}{C}_1& {}^{n+4}{C}_1\\ {}^n{C}_2& {}^{n+2}{C}_2& {}^{n+4}{C}_2\end{array}\right|$ is

(a) 2
(b) 4
(c) 8
(d) n²

Answer 1

$$\begin{gathered} \left|\begin{array}{ccc}1& 1& 1\\ {}^n{C}_1& {}^{n+2}{C}_1& {}^{n+4}{C}_1\\ {}^n{C}_2& {}^{n+2}{C}_2& {}^{n+4}{C}_2\end{array}\right| \\ =\left|\begin{array}{ccc}1& 1& 1\\ n& n+2& n+4\\ \frac{n\left(n-1\right)}{2}& \frac{\left(n+2\right)\left(n+1\right)}{2}& \frac{\left(n+4\right)\left(n+3\right)}{2}\end{array}\right| \\ =\left|\begin{array}{ccc}1& 0& 0\\ n& 2& 4\\ \frac{n\left(n-1\right)}{2}& \frac{4n+2}{2}& \frac{8n+12}{2}\end{array}\right|\left[\mathrm{Applying}{C}_2\to C_2-C_1\mathrm{and}{C}_3\to C_3-C_1\right] \\ =\left|\begin{array}{ccc}1& 0& 0\\ n& 2& 4\\ \frac{n\left(n-1\right)}{2}& \left(2n+1\right)& \left(4n+6\right)\end{array}\right| \\ =8n+12-8n-4 \\ =8 \end{gathered}$$

Hence, the correct option is (c).