Question

Let $\alpha$ and $\beta$ be two real numbers such that $\alpha +\beta =1$and $\alpha \beta =–1$. Let$P_n={\left(\alpha \right)}^n+{\left(\beta \right)}^n,P_{n-1}=11$ and $P_{n+1}=29$ for some integer $n\ge 1$Then, the value of ${P_n}^2$ is______________.

Answer 1

Step 1: Finding ${\alpha }^{n+1}$ and ${\beta }^{n+1}$

Sum of roots of $x^2–x–1=0...\left(i\right)$

$\Rightarrow \alpha$$+\beta =1$

Product of roots

$\Rightarrow ɑ\beta =–1$

Since $\alpha$ be the root of $\left(i\right)$

$$\begin{gathered} \Rightarrow {\alpha }^2–\alpha –1=0 \\ \Rightarrow {\alpha }^2=\alpha +1 \\ \Rightarrow {\alpha }^2={\alpha }^1+{\alpha }^0 \\ \Rightarrow {\alpha }^{1+1}={\alpha }^1+{\alpha }^{1-1} \\ \Rightarrow {\alpha }^{n+1}={\alpha }^n+{\alpha }^{n-1}...\left(ii\right) \end{gathered}$$

Similarly ${\beta }^{n+1}={\beta }^n+{\beta }^{n-1}...\left(iii\right)$

Step 2: Finding the value of ${P_n}^2$

Adding equations $\left(ii\right)$and $\left(iii\right)$

$$\begin{gathered} \Rightarrow {\alpha }^{n+1}+{\beta }^{n+1}={\left(\alpha \right)}^n+{\left(\beta \right)}^n+[{\left(\alpha \right)}^{n+1}+{\left(\beta \right)}^{n+1}] \\ \Rightarrow P_{n+1}=P_n+P_{n-1}\left[\because P_n={\left(\alpha \right)}^n+{\left(\beta \right)}^n\right] \\ \Rightarrow 29=P_n+11\left[\because P_{n-1}=11\&P_{n+1}=29\right] \\ \Rightarrow P_n=29-11 \\ \Rightarrow P_n=18 \\ \Rightarrow {P_n}^2=324 \end{gathered}$$

Hence, the value of ${P_n}^2=324$.