Question

The first four terms of a sequence are given by $T_1=0,T_2=1,T_3=1,T_4=2$. The general term is given by $T_n=A{\alpha }^{n-1}+B{\beta }^{n-1}$ where $A,B,\alpha ,\beta$ are independent of $n$ and $A$ is positive.
The quadratic equation whose roots are $\alpha$ and $\beta$ is given by

• A. $x^2-2x-1=0$
• B. $x^2-2x-2=0$
• C. $x^2-x-1=0$
• D. None

Answer 1

Answer: (C)

We have
$T_1=0...(i)$
$\Rightarrow A+B=0$
$T_2=1...(ii)$
$\Rightarrow A\alpha +B\beta =1$
$T_3=A{\alpha }^2+B{\beta }^2...(iii)$
$=1$
$T_4=A{\alpha }^3+B{\beta }^3=2...(iv)$
From (i), we get $A=-B$

$\therefore \alpha$ and $\beta$ are roots of the equation $x^2-x-1=0$
$\Rightarrow \alpha =\frac{\sqrt{5}+1}{2}$ and $\beta =\frac{1-\sqrt{5}}{2}$
As $(\alpha -\beta )=\sqrt{5}$
so from $A(\alpha -\beta )=1$,
we get $A=\frac{1}{\sqrt{5}}$ and $B=\frac{-1}{\sqrt{5}}($ As $B=-A)$
Hence $5\left(A^2+B^2\right)=2$