$\left\{ a_{n} \right\}\ and\ \{ b_{n}\}$ be two sequences given by $a_{n} = (x)^{\frac{1}{2^{n}}} + (y)^{\frac{1}{2^{n}}}$ and $b_{n} = (x)^{\frac{1}{2^{n}}} - (y)^{\frac{1}{2^{n}}}$ for all $n \in N,$ then $a_{1}a_{2}a_{3}\ldots a_{n}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
$a_{1}a_{2}\ldots a_{n} = b_{n}\left( \frac{a_{1}a_{2}\ldots a_{n}}{b_{n}} \right)$
$= a_{n}b_{n}\left( \frac{a_{1}a_{2}\ldots a_{n - 1}}{b_{n}} \right)$
$= \left\lbrack (x)^{\frac{1}{2^{n - 1}}} - (y)^{\frac{1}{2^{n - 1}}} \right\rbrack\left( \frac{a_{1}a_{2}\ldots a_{n - 1}}{b_{n}} \right)$
$= b_{n - 1} \bullet a_{n - 1}\left( \frac{a_{1}a_{2}\ldots a_{n - 2}}{b_{n}} \right) = ... = \frac{x - y}{b_{n}}$

Q. $\left\{ a_{n} \right\}\ and\ \{ b_{n}\}$ be two sequences given by $a_{n} = (x)^{\frac{1}{2^{n}}} + (y)^{\frac{1}{2^{n}}}$ and $b_{n} = (x)^{\frac{1}{2^{n}}} - (y)^{\frac{1}{2^{n}}}$ for all $n \in N,$ then $a_{1}a_{2}a_{3}\ldots a_{n}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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