If $$\alpha ,\beta \ne 0,$$ and $$f\left( n \right) = {\alpha ^n} + {\beta ^n}$$ and
$$$\left| {\matrix{
3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr
{1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr
{1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr
} } \right|$$$
$$ = K{\left( {1 - \alpha } \right)^2}{\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2},$$ then $$K$$ is equal to :
D
$${1 \over {\alpha \beta }}$$
CHECK ANSWER
Explanation Consider
$$\left| {\matrix{
3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr
{1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr
{1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr
} } \right|$$
$$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{
{1 + 1 + 1} & {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} \cr
{1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} \cr
{1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} & {1 + {\alpha ^4} + {\beta ^4}} \cr
} } \right|$$
$$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{
1 & 1 & 1 \cr
1 & \alpha & \beta \cr
1 & {{\alpha ^2}} & {{\beta ^2}} \cr
} } \right| \times \left| {\matrix{
1 & 1 & 1 \cr
1 & \alpha & \beta \cr
1 & {{\alpha ^2}} & {{\beta ^2}} \cr
} } \right|$$
$$\,\,\,\,\,\,\,\,\,\,\, = {\left| {\matrix{
1 & 1 & 1 \cr
1 & \alpha & \beta \cr
1 & {{\alpha ^2}} & {{\beta ^2}} \cr
} } \right|^2}$$
$$\,\,\,\,\,\,\,\,\,\,\, = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2}$$
So, $$k=1$$