In the sequence an, the n-th term is defined as $a_{n}=\left(a_{n-1}-1\right)^{2}$. If $a_{3}=64$, then what is the vale of $a_{1}$?

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Accepted Answer

সঠিক উত্তর: । ব্যাখ্যা: $a_{3}=\left(a_{2}-1\right)^{2} \Rightarrow 64=a_{2}-1 \Rightarrow a_{2}=9$. একইভাবে, $a_{2}=\left(a_{1}-1\right)^{2} \Rightarrow 9=\left(a_{1}-1\right)^{2} \Rightarrow a_{1}=4$ or $a_{1}=-2$. Since the terms are squares, $a_{1}$ must be a positive integer, so $a_{1}=4$.

In the sequence an, the n-th term is defined as \(a_{n}=\left(a_{n-1}-1\right)^{2}\). If \(a_{3}=64\), then what is the vale of \(a_{1}\)?

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