In a sequence {an};a1=2;an+1=1−1an for n≥1. Let Pn be the product of its first ‘n’ terms; then the value of P2023 is _________

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In a sequence ${a_{n}}; a_{1} = 2; a_{n + 1} = 1 - \frac{1}{a_{n}}$ for $n \geq 1$ . Let $P_{n}$ be the product of its first ‘n’ terms; then the value of $P_{2023}$ is _________

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answer is 2.

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Detailed Solution

Since $a_{n + 2} = 1 - \frac{1}{a_{n + 1}} = 1 - \frac{a_{n}}{a_{n - 1}} = \frac{- 1}{a_{n - 1}}$
It follows that $a_{n + 3} = 1 - \frac{1}{a_{n + 2}} = 1 + a_{n - 1} = a_{n}$
So ${a_{n}}$ is a periodic sequence with period ‘3’ since
$a_{1} = 2, a_{2} = 1 - \frac{1}{2} = \frac{1}{2}, a_{3} = 1 - 2 = - 1$
$a_{1} a_{2} a_{3} = - 1$
$P_{2023} = a_{1} {(a_{1} a_{2} a_{3})}^{674} = 2$