If and be two sequences given by a(n) = (x)^(1/2^n) + (y)^(1/2^n) b(n) = (x)^(1/2^n) - (y)^(1/2^n) for all natural n Then a(1)*a(2)*a(3)*...*a(n) = 1. x - y 2. (x + y)/b(n) 3. (x - y)/b(n) 4. xy/b(n)

Dear Student,

Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{a}\left(\mathrm{n}\right)={\mathrm{x}}^{\frac{1}{2^{\mathrm{n}}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}}}} \\ \mathrm{b}\left(\mathrm{n}\right)={\mathrm{x}}^{\frac{1}{2^{\mathrm{n}}}}-{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}}}} \\ \mathrm{a}\left(1\right)\mathrm{a}\left(2\right)\mathrm{a}\left(3\right)....\mathrm{a}\left(\mathrm{n}\right) \\ =\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2^2}}+{\mathrm{y}}^{\frac{1}{2^2}}\right)\left({\mathrm{x}}^{\frac{1}{2^3}}+{\mathrm{y}}^{\frac{1}{2^3}}\right).....\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}-1}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}-1}}}\right)\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}}}}\right) \\ =\frac{\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2^2}}+{\mathrm{y}}^{\frac{1}{2^2}}\right)\left({\mathrm{x}}^{\frac{1}{2^3}}+{\mathrm{y}}^{\frac{1}{2^3}}\right).....\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}-1}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}-1}}}\right)\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}}}}\right)\mathrm{b}\left(\mathrm{n}\right)}{\mathrm{b}\left(\mathrm{n}\right)} \\ =\frac{\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2^2}}+{\mathrm{y}}^{\frac{1}{2^2}}\right)\left({\mathrm{x}}^{\frac{1}{2^3}}+{\mathrm{y}}^{\frac{1}{2^3}}\right).....\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}-1}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}-1}}}\right)\left\{\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}}}}\right)\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}}}}-{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}}}}\right)\right\}}{\mathrm{b}\left(\mathrm{n}\right)} \\ =\frac{\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2^2}}+{\mathrm{y}}^{\frac{1}{2^2}}\right)\left({\mathrm{x}}^{\frac{1}{2^3}}+{\mathrm{y}}^{\frac{1}{2^3}}\right).....\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}-1}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}-1}}}\right)\left\{\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}}}}\right)\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}}}}-{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}}}}\right)\right\}}{\mathrm{b}\left(\mathrm{n}\right)} \\ =\frac{\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2^2}}+{\mathrm{y}}^{\frac{1}{2^2}}\right)\left({\mathrm{x}}^{\frac{1}{2^3}}+{\mathrm{y}}^{\frac{1}{2^3}}\right).....\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}-1}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}-1}}}\right)\left\{\left({\mathrm{x}}^{\frac{2}{2^{\mathrm{n}}}}-{\mathrm{y}}^{\frac{2}{2^{\mathrm{n}}}}\right)\right\}}{\mathrm{b}\left(\mathrm{n}\right)} \left[\mathrm{Use} \left(\mathrm{A}+\mathrm{B}\right)\left(\mathrm{A}-\mathrm{B}\right)={\mathrm{A}}^2-{\mathrm{B}}^2\right] \\ =\frac{\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2^2}}+{\mathrm{y}}^{\frac{1}{2^2}}\right)\left({\mathrm{x}}^{\frac{1}{2^3}}+{\mathrm{y}}^{\frac{1}{2^3}}\right).....\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}-1}}}+{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}-1}}}\right)\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}-1}}}-{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}-1}}}\right)}{\mathrm{b}\left(\mathrm{n}\right)} \\ =\frac{\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2^2}}+{\mathrm{y}}^{\frac{1}{2^2}}\right)\left({\mathrm{x}}^{\frac{1}{2^3}}+{\mathrm{y}}^{\frac{1}{2^3}}\right).....\left({\mathrm{x}}^{\frac{2}{2^{\mathrm{n}-1}}}-{\mathrm{y}}^{\frac{2}{2^{\mathrm{n}-1}}}\right)}{\mathrm{b}\left(\mathrm{n}\right)} \\ =\frac{\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2^2}}+{\mathrm{y}}^{\frac{1}{2^2}}\right)\left({\mathrm{x}}^{\frac{1}{2^3}}+{\mathrm{y}}^{\frac{1}{2^3}}\right).....\left({\mathrm{x}}^{\frac{1}{2^{\mathrm{n}-2}}}-{\mathrm{y}}^{\frac{1}{2^{\mathrm{n}-2}}}\right)}{\mathrm{b}\left(\mathrm{n}\right)} \\ \mathrm{Similarly} \mathrm{on} \mathrm{reducing} \mathrm{we} \mathrm{get} \\ =\frac{\left({\mathrm{x}}^{\frac{1}{2}}+{\mathrm{y}}^{\frac{1}{2}}\right)\left({\mathrm{x}}^{\frac{1}{2}}-{\mathrm{y}}^{\frac{1}{2}}\right)}{\mathrm{b}\left(\mathrm{n}\right)} \\ =\frac{\mathrm{x}-\mathrm{y}}{\mathrm{b}\left(\mathrm{n}\right)} \left[\mathrm{Answer}\right] \end{gathered}$$

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