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If $\alpha ,\beta \ne 0$ and $f\left( n \right) = {\alpha ^n} + {\beta ^n}$ and $\left| {\begin{array}{*{20}{c}}3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}\\{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}\\{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}\end{array}} \right|\; = K{\left( {1 - \alpha } \right)^2}$ ${\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}$ ,then $K=$ . . . . . .
$1$
$-1$
$\alpha \beta $
$\frac{1}{{\alpha \beta }}$
Solution
$f\left( n \right) = {\alpha ^n} + {\beta ^n}$
$\left| {\begin{array}{*{20}{c}}
{1 + 1 + 1}&{1 + \alpha + \beta }&{1 + {\alpha ^2} + {\beta ^2}}\\
{1 + \alpha + \beta }&{1 + {\alpha ^2} + {\beta ^2}}&{1 + {\alpha ^3} + {\beta ^3}}\\
{1 + {\alpha ^2} + {\beta ^2}}&{1 + {\alpha ^3} + {\beta ^3}}&{1 + {\alpha ^4} + {\beta ^4}}
\end{array}} \right|$
$ = \left| {\begin{array}{*{20}{c}}
1&1&1\\
1&\alpha &{{\alpha ^2}}\\
1&\beta &{{\beta ^2}}
\end{array}} \right|$
$ = {\left( {1 – \alpha } \right)^2}{\left( {\alpha – \beta } \right)^2}{\left( {\beta – 1} \right)^2}$
$\Rightarrow \boxed{k = 1}$
