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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______________.

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Solution

## Step 1: Finding ${\alpha }^{n+1}$ and ${\beta }^{n+1}$Sum of roots of ${x}^{2}–x–1=0...\left(i\right)$$⇒\alpha$$+\beta =1$Product of roots $⇒ɑ\beta =–1$Since $\alpha$ be the root of $\left(i\right)$$⇒{\alpha }^{2}–\alpha –1=0\phantom{\rule{0ex}{0ex}}⇒{\alpha }^{2}=\alpha +1\phantom{\rule{0ex}{0ex}}⇒{\alpha }^{2}={\alpha }^{1}+{\alpha }^{0}\phantom{\rule{0ex}{0ex}}⇒{\alpha }^{1+1}={\alpha }^{1}+{\alpha }^{1-1}\phantom{\rule{0ex}{0ex}}⇒{\alpha }^{n+1}={\alpha }^{n}+{\alpha }^{n-1}...\left(ii\right)$Similarly ${\beta }^{n+1}={\beta }^{n}+{\beta }^{n-1}...\left(iii\right)$Step 2: Finding the value of ${{P}_{n}}^{2}$Adding equations $\left(ii\right)$and $\left(iii\right)$$⇒{\alpha }^{n+1}+{\beta }^{n+1}={\left(\alpha \right)}^{n}+{\left(\beta \right)}^{n}+\left[{\left(\alpha \right)}^{n+1}+{\left(\beta \right)}^{n+1}\right]\phantom{\rule{0ex}{0ex}}⇒{P}_{n+1}={P}_{n}+{P}_{n-1}\left[\because {P}_{n}={\left(\alpha \right)}^{n}+{\left(\beta \right)}^{n}\right]\phantom{\rule{0ex}{0ex}}⇒29={P}_{n}+11\left[\because {P}_{n-1}=11&{P}_{n+1}=29\right]\phantom{\rule{0ex}{0ex}}⇒{P}_{n}=29-11\phantom{\rule{0ex}{0ex}}⇒{P}_{n}=18\phantom{\rule{0ex}{0ex}}⇒{{P}_{n}}^{2}=324$Hence, the value of ${{P}_{n}}^{2}=324$.

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