Question

If $ɑ,\beta \ne 0,f\left(n\right)={ɑ}^n+{\beta }^n$ and $\left|\begin{array}{ccc}3& 1+f\left(1\right)& 1+f\left(2\right)\\ 1+f\left(1\right)& 1+f\left(2\right)& 1+f\left(3\right)\\ 1+f\left(2\right)+1+f\left(3\right)& 1+f\left(3\right)& 1+f\left(4\right)\end{array}\right|$ $=k{(1-ɑ)}^2{(1-\beta )}^2{(ɑ-\beta )}^2$, then $k=$

• A. $\alpha \beta$
• B. $\frac{1}{\alpha \beta }$
• C. $1$
• D. $-1$

Answer 1

The correct option is C

$1$

Explanation for the correct option:

Step 1. Find the value of $k$:

Given, $f\left(n\right)={ɑ}^n+{\beta }^n$

$\Rightarrow$ $f\left(1\right)=\alpha +\beta ,f\left(2\right)={\alpha }^2+{\beta }^2,f\left(3\right)={\alpha }^3+{\beta }^3,f\left(4\right)={\alpha }^4+{\beta }^4$

Let $△=\left|\begin{array}{ccc}3& 1+f\left(1\right)& 1+f\left(2\right)\\ 1+f\left(1\right)& 1+f\left(2\right)& 1+f\left(3\right)\\ 1+f\left(2\right)+1+f\left(3\right)& 1+f\left(3\right)& 1+f\left(4\right)\end{array}\right|$

Step 2. Put the values of $f\left(1\right),f\left(2\right),f\left(3\right),f\left(4\right)$ in $△$, we get

$△=\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$

$\begin{array}{rcl}△& =& \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& \beta & {\beta }^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& \beta & {\beta }^2\end{array}\right|\\ & =& {\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& \beta & {\beta }^2\end{array}\right|}^2\end{array}$

$\therefore △={\left(1-\alpha \right)}^2{\left(1-\beta \right)}^2{\left(\alpha -\beta \right)}^2$

Step 3. According to the question,$△=k{\left(1-\alpha \right)}^2{\left(1-\beta \right)}^2{\left(\alpha -\beta \right)}^2$

$\Rightarrow$ ${\left(1-\alpha \right)}^2{\left(1-\beta \right)}^2{\left(\alpha -\beta \right)}^2=k{\left(1-\alpha \right)}^2{\left(1-\beta \right)}^2{\left(\alpha -\beta \right)}^2$

By Comparing L.H.S and R.H.S, we get

$\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, Option ‘C’ is Correct.