Question:
Let \(\alpha\) and \(\beta\) be two real numbers such that \(\alpha+\beta=1\) and \(\alpha\beta=-1\). Let \(p_n=(\alpha)^n+(\beta)^n\), \(p_{n-1}=11\) and \(p_{n+1}=29\) for some integer \(n\ge1\). Then, the value of \(p_n^2\) is ___________.
Let \(\alpha\) and \(\beta\) be two real numbers such that \(\alpha+\beta=1\) and \(\alpha\beta=-1\). Let \(p_n=(\alpha)^n+(\beta)^n\), \(p_{n-1}=11\) and \(p_{n+1}=29\) for some integer \(n\ge1\). Then, the value of \(p_n^2\) is ___________.
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\(\alpha,\beta\) are roots of \(x^2-x-1=0\), so \(p_{n+1}=p_n+p_{n-1}\).
Updated On: Nov 4, 2026
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Correct Answer: 324
Solution and Explanation
\(\alpha,\beta\) are roots of \(x^2-x-1=0\), so \(\alpha^2=\alpha+1\) and \(\beta^2=\beta+1\).
Multiplying by \(\alpha^{n-1}\), \(\beta^{n-1}\) and adding: \(p_{n+1}=p_n+p_{n-1}\).
\(29=p_n+11\Rightarrow p_n=18\), so \(p_n^2=324\).
(Check: the sequence \(p_k\) is \(2,1,3,4,7,11,18,29\), with \(p_{n-1}=11,p_n=18,p_{n+1}=29\).)
Answer: 324
Multiplying by \(\alpha^{n-1}\), \(\beta^{n-1}\) and adding: \(p_{n+1}=p_n+p_{n-1}\).
\(29=p_n+11\Rightarrow p_n=18\), so \(p_n^2=324\).
(Check: the sequence \(p_k\) is \(2,1,3,4,7,11,18,29\), with \(p_{n-1}=11,p_n=18,p_{n+1}=29\).)
Answer: 324
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