MCQ
The value of $\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$ is :
  • A
    $2$
  • B
    $4$
  • $8$
  • D
    $n^2$

Answer

Correct option: C.
$8$
$\begin{vmatrix}1&1&1\\^\text{n}\text{C}_1&^{\text{n}+2}\text{C}_1&^{\text{n}+4}\text{C}_1\\^\text{n}\text{C}_2&^{\text{n}+2}\text{C}_2&^{\text{n}+4}\text{C}_2\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\\text{n}&\text{n}+2&\text{n}+3\\\frac{\text{n}(\text{n}-1)}{2}&\frac{(\text{n}+2)(\text{n}+1)}{2}&\frac{(\text{n}+4)(\text{n}+3)}{2}\end{vmatrix}$
$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&\frac{4\text{n}+2}{2}&\frac{8\text{n}+12}{2}\end{vmatrix}\ [$Applying $C _2 \rightarrow C _2- C _1$ and $\left.C _3 \rightarrow C _3- C _1\right]$
$=\begin{vmatrix}1&0&0\\\text{n}&2&4\\\frac{\text{n}(\text{n}+1)}{2}&(2\text{n}+1)&(4\text{n}+6)\end{vmatrix}$
$=8\text{n}+12-8\text{n}-4$
$=8$
Hence, the correct option is $(c)$

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