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 ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$ and if ${{a}_{4}}=28$, then find the value of $p+2q$?
(a) 7
(b) 21
(c) 14
(d) 12

Answer 1

Hint:We start solving the problem by finding the roots \[\alpha \], $\beta $ of the given quadratic equation using the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We then substitute these obtained roots in ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$ for $n=4$. After making the necessary calculations, we can see that the R.H.S has integer part and irrational part but the L.H.S part only has integer part. We equate irrational part to 0 to find the values of $p$ and $q$. Using these values, we find the value of $p+2q$.

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 find the value of $p+2q$, if ${{a}_{4}}=28$ and p, q are integers.
Let us first find the roots \[\alpha \], $\beta $ of the equation ${{x}^{2}}-x-1=0$.
We know that the roots of the equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
So, the roots of ${{x}^{2}}-x-1=0$ are $\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)}$.
$\Rightarrow \dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)}=\dfrac{1\pm \sqrt{1+4}}{2}$.
$\Rightarrow \dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)}=\dfrac{1\pm \sqrt{5}}{2}$.
Let $\alpha =\dfrac{1+\sqrt{5}}{2}$ and $\beta =\dfrac{1-\sqrt{5}}{2}$.
According to the problem, we have ${{a}_{4}}=28$ and ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$.
So, we get $28=p{{\alpha }^{4}}+q{{\beta }^{4}}$.
$\Rightarrow 28=p{{\left( \dfrac{1+\sqrt{5}}{2} \right)}^{4}}+q{{\left( \dfrac{1-\sqrt{5}}{2} \right)}^{4}}$.
$\Rightarrow 28=p\dfrac{{{\left( 1+\sqrt{5} \right)}^{4}}}{{{2}^{4}}}+q\dfrac{{{\left( 1-\sqrt{5} \right)}^{4}}}{{{2}^{4}}}$.
We know that ${{\left( 1+x \right)}^{4}}=1+4x+6{{x}^{2}}+4{{x}^{3}}+{{x}^{4}}$.
$\Rightarrow 28=p\dfrac{\left( 1+4\left( \sqrt{5} \right)+6{{\left( \sqrt{5} \right)}^{2}}+4{{\left( \sqrt{5} \right)}^{3}}+{{\left( \sqrt{5} \right)}^{4}} \right)}{16}+q\dfrac{\left( 1+4\left( -\sqrt{5} \right)+6{{\left( -\sqrt{5} \right)}^{2}}+4{{\left( -\sqrt{5} \right)}^{3}}+{{\left( -\sqrt{5} \right)}^{4}} \right)}{16}$.
$\Rightarrow 28=p\dfrac{\left( 1+4\sqrt{5}+6\left( 5 \right)+4\left( 5\sqrt{5} \right)+\left( 25 \right) \right)}{16}+q\dfrac{\left( 1-4\sqrt{5}+6\left( 5 \right)+4\left( -5\sqrt{5} \right)+\left( 25 \right) \right)}{16}$.
$\Rightarrow 28=p\dfrac{\left( 26+4\sqrt{5}+30+20\sqrt{5} \right)}{16}+q\dfrac{\left( 26-4\sqrt{5}+30-20\sqrt{5} \right)}{16}$.
$\Rightarrow 28=p\dfrac{\left( 56+24\sqrt{5} \right)}{16}+q\dfrac{\left( 56-24\sqrt{5} \right)}{16}$.
$\Rightarrow 28=\dfrac{56\left( p+q \right)}{16}+\dfrac{24\sqrt{5}\left( p-q \right)}{16}$.
$\Rightarrow 28=\dfrac{7\left( p+q \right)}{2}+\dfrac{3\sqrt{5}\left( p-q \right)}{2}$.
$\Rightarrow 28\times 2=7\left( p+q \right)+3\sqrt{5}\left( p-q \right)$.
$\Rightarrow 56=7\left( p+q \right)+3\sqrt{5}\left( p-q \right)$ ---(1).
According to the problem, it is given that p and q are integers. We can see that $3\sqrt{5}$ is an irrational number. We know that the subtraction or addition of two integers gives integers.
So, we get $\left( p-q \right)$ as an integer. We know that the multiplication of irrational numbers and an integer gives an irrational numbers. So, we get $3\sqrt{5}\left( p-q \right)$ as an irrational number. But we have the only integer in L.H.S (Left Hand Side). So, there should not be any irrational numbers present in the R.H.S (Right Hand Side). This happens only when $p-q=0$.
So, we get $p=q$. We use this result in equation (1).
$\Rightarrow 56=7\left( p+p \right)+3\sqrt{5}\left( p-p \right)$.
$\Rightarrow 56=7\left( 2p \right)+3\sqrt{5}\left( 0 \right)$.
$\Rightarrow 56=14p$.
$\Rightarrow p=\dfrac{56}{14}$.
$\Rightarrow p=4$.
Since $p=q$, we get $q=4$.
Now, let us find the value of $p+2q$.
So, we have $p+2q=4+2\left( 4 \right)$.
$\Rightarrow p+2q=4+8$.
$\Rightarrow p+2q=12$.
We have found the value of $p+2q$ as 12.
∴ The value of $p+2q$ is 12.
The correct option for the given problem is (d).

Note:
 We should know that sum or addition or multiplication or division of integer with an irrational number gives results as an irrational number. Since p and q are clearly mentioned as integers in the problem, we take the irrational part as zero. We can also find the general form of ${{a}_{n}}=p{{\alpha }^{n}}+q{{\beta }^{n}}$ in terms of ${{a}_{n-1}}$ and ${{a}_{n-2}}$ which will reduce our calculation time. We should not calculation mistakes while solving this problem.