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Question
The value of \[\begin{vmatrix}1 & 1 & 1 \\ {}^n C_1 & {}^{n + 2} C_1 & {}^{n + 4} C_1 \\ {}^n C_2 & {}^{n + 2} C_2 & {}^{n + 4} C_2\end{vmatrix}\] is
Options
2
4
8
n2
Solution
\[\begin{vmatrix}1 & 1 & 1 \\ {}^n C_1 & {}^{n + 2} C_1 & {}^{n + 4} C_1 \\ {}^n C_2 & {}^{n + 2} C_2 & {}^{n + 4} C_2\end{vmatrix}\]
\[ = \begin{vmatrix}1 & 1 & 1 \\ n & n + 2 & n + 4 \\ \frac{n\left( n - 1 \right)}{2} & \frac{\left( n + 2 \right)\left( n + 1 \right)}{2} & \frac{\left( n + 4 \right)\left( n + 3 \right)}{2}\end{vmatrix}\]
\[ = \begin{vmatrix}1 & 0 & 0 \\ n & 2 & 4 \\ \frac{n\left( n - 1 \right)}{2} & \frac{4n + 2}{2} & \frac{8n + 12}{2}\end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_1\text{ and }C_3 \to C_3 - C_1 \right]\]
\[ = \begin{vmatrix}1 & 0 & 0 \\ n & 2 & 4 \\ \frac{n\left( n - 1 \right)}{2} & \left( 2n + 1 \right) & \left( 4n + 6 \right)\end{vmatrix}\]
\[ = 8n + 12 - 8n - 4\]
\[ = 8\]
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