Question

If \(\) and \(\) are two sequences given by $a_{1}=2^{1 / 2}+3^{1 / 2}$ \(a_{2}=2^{14}+3^{1 / 4}\) \(a_{3}=2^{1 / 8}+3^{1 / 8}\) and \(b_{1}=2^{1 / 2}-3^{1 / 2}\) \(b_{2}=2^{1 / 4}-3^{1 / 4}\) \(b_{3}=2^{1 / 8}-3^{1 / 8}\) \(\ldots \ldots \ldots \ldots\) Then $\mathrm{a}_{1} \mathrm{a}_{2} \mathrm{a}_{3} \ldots \mathrm{a}_{\mathrm{n}}$ equals (a) \(\mathrm{b}_{1} \mathrm{~b}_{2} \ldots \mathrm{b}_{=}\) (b) \(\frac{1}{b_{1} b_{2} \ldots b_{n}}\) (c) \(\frac{1}{b_{\mathrm{n}}}\) (d) \(\frac{-1}{\mathrm{~b}_{\mathrm{n}}}\)

Short answer

Answer: The relationship between the product of the terms of sequence \(\) and the terms of sequence \(\) is given by \(a_1a_2a_3\ldots a_n = \frac{1}{b_1b_2\ldots b_n}\).

Step-by-step solution

Calculate products for \(\) and \(\)

First, let's multiply the terms \(a_1a_2\) and \(b_1b_2\). This will give us an idea of the relationship between the two sequences. - \(a_1a_2 = (2^{1/2}+3^{1/2})(2^{1/4}+3^{1/4})\) - \(b_1b_2 = (2^{1/2}-3^{1/2})(2^{1/4}-3^{1/4})\)

Find the relationship

Now let's use difference of squares to simplify the expressions: - \(a_1a_2 = (2^{1/2}+3^{1/2})(2^{1/4}+3^{1/4}) = (2^{1/2}+3^{1/2})^2 - (2^{1/4}-3^{1/4})^2\) - \(b_1b_2 = (2^{1/2}-3^{1/2})(2^{1/4}-3^{1/4}) = (2^{1/2}+3^{1/2})(2^{1/4}+3^{1/4}) - (2^{1/2}-3^{1/2})^2\) We can see a relationship between \(a_1a_2\) and \(b_1b_2\): - \(a_1a_2 = (2^{1/2}+3^{1/2})^2 - (2^{1/4}-3^{1/4})^2 = (2^{1/2}+3^{1/2})^2 - b_1b_2\)

Determine the correct relationship

Now let us check our possible relations by multiplying \(a_n\) terms with suitable \(b_n\) terms to see if any of the given options match our findings: - Option (a): \(a_1a_2\ldots a_n \neq b_1b_2\ldots b_n\) - Option (b): \(a_1a_2\ldots a_n = \frac{1}{b_1b_2\ldots b_n} \implies a_1a_2 = \frac{1}{b_1b_2} \implies a_1a_2 = \frac{1}{(2^{1/2}+3^{1/2})^2 - b_1b_2}\) - Option (c): \(a_1a_2\ldots a_n \neq \frac{1}{b_n}\) - Option (d): \(a_1a_2\ldots a_n \neq \frac{-1}{b_n}\) Comparing our result from Step 2 with the possible relations, we see that Option (b) matches our findings: $$a_1a_2 = \frac{1}{b_1b_2}$$ Hence, the relationship between the two sequences is $$a_1a_2a_3\ldots a_n= \frac{1}{b_1b_2\ldots b_n}$$