Question

The value of $({\alpha }^2+{\beta }^2+\alpha \beta )$ is equal to

• A. $1$
• B. $2$
• C. $5$
• D. $4$

Answer 1

Answer: (B)

$2$

Given that, $T_n=A{\alpha }^{n-1}+B{\beta }^{n-1}$
$\because T_1=0$
$\therefore A+B=0$
$\Rightarrow A=-B$ ....(1)
$\because T_2=1$
$\therefore A\alpha +B\beta =1$
$\Rightarrow A(\alpha -\beta )=1$ ....(2) [ From (1) ]
$\because T_4=2$
$\therefore A{\alpha }^3+B{\beta }^3=2$
$\Rightarrow A({\alpha }^3-{\beta }^3)=2$ ....[ From (1) ]

$\Rightarrow A(\alpha -\beta )({\alpha }^2+{\beta }^2+\alpha \beta )=2$
$\Rightarrow {\alpha }^2+{\beta }^2+\alpha \beta =2$ ....[ From (2) ]
Ans: B