Question

Let p, q be the integers and let \[\alpha \], $\beta $ be the roots of the equation ${{x}^{2}}-x-1=0$, where $\alpha \ne \beta $. For n = 0, 1, 2, ……, Let ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$, then ${{a}_{12}}$=?
(a) ${{a}_{11}}+{{a}_{10}}$,
(b) ${{a}_{11}}-{{a}_{10}}$,
(c) ${{a}_{11}}+2{{a}_{10}}$,
(d) $2{{a}_{11}}+{{a}_{10}}$.

Answer 1

Hint:We start solving the problem by modifying the given quadratic equation. We multiply both sides of the modified equation with ${{x}^{10}}$ and substitute $\alpha $ in the obtained equation and multiply it with p to get our first condition. We then substitute $\beta $ in the obtained equation and multiply it with q to get our second condition. We then add both the conditions and compare it with ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$ to get the desired result for the problem.

Complete step by step answer:
According to the problem, we are given that \[\alpha \], $\beta $ be the roots of the equation ${{x}^{2}}-x-1=0$ and ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$, for n = 0, 1, 2, ……. We need to convert ${{a}_{12}}$ in terms of ${{a}_{11}}$ and ${{a}_{10}}$.
We have the quadratic equation ${{x}^{2}}-x-1=0$. Let us re-modify this equation.
So, we get the quadratic equation as ${{x}^{2}}=x+1$.
Now, let us multiply both sides of the quadratic equation with ${{x}^{10}}$.
So, we get ${{x}^{10}}\left( {{x}^{2}} \right)={{x}^{10}}\left( x+1 \right)$.
$\Rightarrow {{x}^{12}}={{x}^{11}}+{{x}^{10}}$ ---(1).
Since, \[\alpha \], $\beta $ be the roots of the equation ${{x}^{2}}-x-1=0$. This also becomes the roots of ${{x}^{12}}={{x}^{11}}+{{x}^{10}}$ as we multiplied ${{x}^{10}}$ to the quadratic equation ${{x}^{2}}-x-1=0$.
So, let us substitute \[\alpha \] in equation (1).
\[\Rightarrow {{\alpha }^{12}}={{\alpha }^{11}}+{{\alpha }^{10}}\]. We now multiply both sides with p.
\[\Rightarrow p{{\alpha }^{12}}=p{{\alpha }^{11}}+p{{\alpha }^{10}}\] ---(2).
Now, let us substitute $\beta $ in equation (1).
\[\Rightarrow {{\beta }^{12}}={{\beta }^{11}}+{{\beta }^{10}}\]. We now multiply both sides with q.
\[\Rightarrow q{{\beta }^{12}}=q{{\beta }^{11}}+q{{\beta }^{10}}\] ---(3).
Let us add equations (2) and (3).
\[\Rightarrow p{{\alpha }^{12}}+q{{\beta }^{12}}=p{{\alpha }^{11}}+p{{\alpha }^{10}}+q{{\beta }^{11}}+q{{\beta }^{10}}\].
\[\Rightarrow p{{\alpha }^{12}}+q{{\beta }^{12}}=\left( p{{\alpha }^{11}}+q{{\beta }^{11}} \right)+\left( p{{\alpha }^{10}}+q{{\beta }^{10}} \right)\].
We can see that the terms resemble ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$.
So, we get \[{{a}_{12}}={{a}_{11}}+{{a}_{10}}\].
We have converted ${{a}_{12}}$ in terms of \[{{a}_{11}}\] and \[{{a}_{10}}\] as \[{{a}_{12}}={{a}_{11}}+{{a}_{10}}\].
The correct option for the given problem is (a).

Note:
We can also find the values of $\alpha $, $\beta $ using the fact that the roots of the equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We then substitute this in ${{a}_{12}}$, \[{{a}_{11}}\] and ${{a}_{10}}$ to get the result. Here we can see that this approach requires huge amount of calculation and there is a high chance of making mistakes. We should not confuse with the given notation ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$, while solving this problem.