Question

The value of ${\sum }_{r=2}^n{(-2)}^r\left|\begin{array}{ccc}{{}^{n-2}C}_{r-2}& {{}^{n-2}C}_{r-1}& {{}^{n-2}C}_r\\ -3& 1& 1\\ 2& -1& 0\end{array}\right|(n>2)$

• A. $2n-1+{(-1)}^n$
• B. $2n-1+{(-1)}^{n-1}$
• C. $2n-3+{(-1)}^n$
• D. none of these

Answer 1

Answer: (A)

$2n-1+{(-1)}^n$

$\Delta ={\sum }_{r=2}^n{(-2)}^r\left|\begin{array}{ccc}{{}^{n-2}C}_{r-2}& {{}^{n-2}C}_{r-1}& {{}^{n-2}C}_r\\ -3& 1& 1\\ 2& -1& 0\end{array}\right|$
On expanding , we get
$={\sum }_{r=2}^n(-2{)}^r{[}^{n-2}{C}_{r-2}+2^{n-2}{C}_{r-1}{+}^{n-2}{C}_r]$
$={\sum }_{r=2}^n(-2{)}^r{[}^{n-2}{C}_{r-2}{+}^{n-2}{C}_{r-1}{+}^{n-2}{C}_{r-1}{+}^{n-2}{C}_r]$
$={\sum }_{r=2}^n(-2{)}^r{[}^{n-1}{C}_{r-1}{+}^{n-1}{C}_r]$
$={\sum }_{r=2}^n(-2{)}^r{[}^n{C}_r]$
$\Delta =(-2{)}^2{}^n{C}_2+(-2{)}^3{}^n{C}_3+......+(-2{)}^n{}^n{C}_n$ .....(2)
Now, we will expand $(1-2{)}^n$
$(1-2{)}^n{=}^n{C}_0{+}^n{C}_1(-2\left){+}^n{C}_2\right(-2{)}^4{+}^n{C}_3(-2{)}^3+....{.}^n{C}_n(-2{)}^n$
$\Rightarrow (-1{)}^n-1+2n{=}^n{C}_2(-2{)}^4{+}^n{C}_3(-2{)}^3+....{.}^n{C}_n(-2{)}^n$
Using this in (1), we get
$\Delta =2n-1+(-1{)}^n$