For positive x > y , define a n = x 1/ 2 n + y 1/ 2 n and b n = x 1/ 2 n − y 1/ 2 n for n ≥ 1 . Find a 1 a 2 ⋯ a n .

a n b n = x 1/ 2 n − 1 − y 1/ 2 n − 1 = b n − 1 for n ≥ 2 , and a 1 b 1 = x − y . Multiplying and cancelling gives ( a 1 ⋯ a n ) b n = x − y .

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MathematicsSequence & SeriesSpecial sequences/seriesMedium2 minQB

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For positive $x > y$ , define $a_{n} = x^{1/ 2^{n}} + y^{1/ 2^{n}}$ and $b_{n} = x^{1/ 2^{n}} - y^{1/ 2^{n}}$ for $n \geq 1$ . Find $a_{1} a_{2} \dots a_{n}$ .

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Answer:

Solution:

$a_{n} b_{n} = x^{1/ 2^{n - 1}} - y^{1/ 2^{n - 1}} = b_{n - 1}$ for $n \geq 2$ , and $a_{1} b_{1} = x - y$ . Multiplying and cancelling gives $(a_{1} \dots a_{n}) b_{n} = x - y$ .

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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