A sequence has \( n \)th term \( a_{n}=\cos \left(90 n^{\circ}\right), n \geqslant 1 \). a Find the order of the sequence. b Find \( \sum_{r=1}^{444} a_{r} \)

To solve the given problem, let's start by analyzing the sequence defined by the \( n \)th term \( a_n = \cos(90n^\circ) \). ### Part a: Find the order of the sequence. We begin by simplifying \( a_n \): \[ a_n = \cos(90n^\circ) \] We know that \( \cos(90^\circ) = 0 \), \( \cos(180^\circ) = -1 \), and \( \cos(270^\circ) = 0 \), while \( \cos(360^\circ) = 1 \). Thus, the cosine function cycles every 360 degrees. Notably, \( 90n^\circ \) results in: - For \( n = 1 \): \( a_1 = \cos(90^\circ) = 0 \) - For \( n = 2 \): \( a_2 = \cos(180^\circ) = -1 \) - For \( n = 3 \): \( a_3 = \cos(270^\circ) = 0 \) - For \( n = 4 \): \( a_4 = \cos(360^\circ) = 1 \) - For \( n = 5 \): \( a_5 = \cos(450^\circ) = 0 \) - For \( n = 6 \): \( a_6 = \cos(540^\circ) = -1 \) From this, we see that: - \( a_n = 0 \) if \( n \equiv 1 \, (\text{mod } 4) \) - \( a_n = -1 \) if \( n \equiv 2 \, (\text{mod } 4) \) - \( a_n = 0 \) if \( n \equiv 3 \, (\text{mod } 4) \) - \( a_n = 1 \) if \( n \equiv 0 \, (\text{mod } 4) \) Thus, the sequence oscillates between 0, -1, 0, and 1 every four terms. Therefore, we can determine that the order of the sequence is periodic with a period of 4. ### Part b: Find \( \sum_{r=1}^{444} a_{r} \). To find \( \sum_{r=1}^{444} a_{r} \), we will break it down based on the periodic pattern we found. In one complete cycle (from \( r = 1 \) to \( r = 4 \)), we have: \[ a_1 + a_2 + a_3 + a_4 = 0 + (-1) + 0 + 1 = 0 \] Since \( a_n \) has a period of 4, the sum of every complete set (4 terms) is 0. We need to determine how many complete cycles fit into 444 terms and if there are any extra terms: \[ \text{Number of complete cycles} = \left\lfloor \frac{444}{4} \right\rfloor = 111 \] This means there are 111 complete cycles contributing \( 111 \times 0 = 0 \) to the sum. Next, we check for any remaining terms: \[ \text{Remaining terms} = 444 \mod 4 = 0 \quad (No additional terms) \] Thus, the sum of all terms from \( r = 1 \) to \( r = 444 \) is: \[ \sum_{r=1}^{444} a_{r} = 111 \cdot 0 + 0 = 0. \] ### Final Answer: a. The order of the sequence is 4. b. \( \sum_{r=1}^{444} a_{r} = 0 \).