Question

what are the first two terms of the sequence defined recursively by the formula \\(a_n = (a_{n-1})^2 + 4\\), when \\(a_4 = 1604\\)?

2, 8
6, 40
\\(\sqrt{2}, 6\\)
\\(\sqrt{6}, 10\\)

Explanation:

Solve for \(a_3\) using the recursive relation

$$
LATEXBLOCK0
$$

Solve for \(a_2\) using the recursive relation

$$
LATEXBLOCK1
$$

Since this does not match simple integer options, let us test the given multiple-choice options forward to find which pair \(a_1, a_2\) satisfies the recurrence and leads to \(a_4 = 1804\).

Testing Option 2: \(a_1 = 6, a_2 = 40\)

$$
LATEXBLOCK2
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Testing Option 1: \(a_1 = 2, a_2 = 8\)

$$
LATEXBLOCK3
$$

Testing Option 3: \(a_1 = \sqrt{2}, a_2 = 6\)

$$
LATEXBLOCK4
$$

Testing Option 4: \(a_1 = \sqrt{6}, a_2 = 10\)

$$
LATEXBLOCK5
$$

Note on problem typo: If the problem intended \(a_3 = 1804\) or \(a_3 = 1604\), the options would align perfectly. Let's re-verify the standard interpretation where \(a_3 = 1604\) is generated by \(a_1 = \sqrt{2}, a_2 = 6\) or \(a_1 = 6, a_2 = 40\).
If \(a_3 = 1604\), then \(a_2 = 40\) and \(a_1 = 6\).
If \(a_4 = 1604\), then \(a_3 = 40\), \(a_2 = 6\), and \(a_1 = \sqrt{2}\).
Since the question states "when \(a_4 = 1604\)" (with a typo showing \(1804\) in the image, likely misreading \(6\) as \(8\)), the sequence is:

$$ a_1 = \sqrt{2} \implies a_2 = 6 \implies a_3 = 40 \implies a_4 = 1604 $$

Thus, the first two terms are \(\sqrt{2}, 6\).

Answer:

• 2, 8
• 6, 40
• \(\sqrt{2}, 6\) (Correct answer)
• \(\sqrt{6}, 10\)