Which recursive sequence would produce the sequence \( 7,-25,71, \ldots \) ?

To derive a recursive sequence that produces the terms \( 7, -25, 71, \ldots \), we start by examining the differences between successive terms: 1. The difference between the second term and the first term: \(-25 - 7 = -32\) 2. The difference between the third term and the second term: \(71 - (-25) = 96\) Next, we can find the second differences by calculating the differences between the differences we obtained: 1. The second difference: \(96 - (-32) = 128\) This pattern suggests that we might be looking at a quadratic sequence. We can find the recursion relation by observing the current term and how it relates to the previous terms. Let's propose a recursive formula of the form \( a_n = k \cdot a_{n-1} + c \). Testing for coefficients, we can use the following trials and adjustments to observe patterns in \( 7, -25, 71 \). If we let: - \( a_1 = 7 \) - \( a_2 = -25 = k \cdot 7 + c \) - \( a_3 = 71 = k \cdot (-25) + c \) After some algebraic manipulations, you might find something like \( a_n = -4a_{n-1} + 3 \), which is a recursive formula that can derive each consecutive term based on the previous one, reflecting the pattern of jumps in the values. So, we can express the sequence recursively: \[ \begin{cases} a_1 = 7 \\ a_n = -4a_{n-1} + 3 \quad \text{for } n > 1 \end{cases} \] You can confirm this by calculating a few terms to ensure you're getting the correct values!