卷別基礎數學選擇國立中興大學 2025 · 考試 · 第 13 題 考科離散數學(2 points) Solve the recurrence equation an=an−1+2an−2a_n = a_{n-1} + 2a_{n-2}an=an−1+2an−2, with a0=0a_0 = 0a0=0 and a1=1a_1 = 1a1=1, and you can get an=p[qn+1+rn]a_n = p[q^{n+1} + r^n]an=p[qn+1+rn], n≥0n \ge 0n≥0. What are the values of (p,q,r)(p, q, r)(p,q,r)?(A)(3,2,1)(3, 2, 1)(3,2,1)(B)(1/3,−1,−2)(1/3, -1, -2)(1/3,−1,−2)(C)(1/3,1,2)(1/3, 1, 2)(1/3,1,2)(D)(3,−1,2)(3, -1, 2)(3,−1,2)(E)(1/3,−1,2)(1/3, -1, 2)(1/3,−1,2)標準解析AI 已複查顯示標準解析 上一題第 12 題第 14 題下一題
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