Word Count : 506

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We are given the recurrence relation: [ u_n + \alpha u_{n-1} + \beta u_{n-2} = f(n), \quad n \ge 2 ] and a solution: [ u_n = 1 - 2n + 3 \cdot 2^n ] We need to find (\alpha), (\beta), and (f(n)). --- Step 1: Express (u_{n-1}) and (u_{n-2}) [ u_{n-1} = 1 - 2(n-1) + 3 \cdot 2^{n-1} = 1 - 2n + 2 + \frac{3}{2} \cdot 2^n = 3 - 2n + \frac{3}{2} \cdot 2^n ] [ u_{n-2} = 1 - 2(n-2) + 3 \cdot 2^{n-2} = 1 - 2n + 4 + \frac{3}{4} \cdot 2^n = 5 - 2n + \frac{3}{4} \cdot 2^n ] --- Step 2: Substitute into the recurrence [ u_n + \alpha u_{n-1} + \beta u_{n-2} = (1 - ___ _______ _______ __________ ____ ______ _________ _______ _______ ____ _________.
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