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國立臺灣大學 2022 · 考試 · 第 2 題

考科離散數學

(10%) Which solves an=an−1+6an−2a_n = a_{n-1} + 6a_{n-2} for ana_n in terms of a0=Aa_0 = A and a1=Ba_1 = B:

  • (A)
    15[(−3)n(2A−B)+2n(3A+B)]\frac{1}{5}\left[(-3)^n(2A-B) + 2^n(3A+B)\right]
  • (B)
    15[(−3)n(2A−B)+2n(3A−B)]\frac{1}{5}\left[(-3)^n(2A-B) + 2^n(3A-B)\right]
  • (C)
    15[(−2)n(3A−B)+3n(2A+B)]\frac{1}{5}\left[(-2)^n(3A-B) + 3^n(2A+B)\right]
  • (D)
    15[(−2)n(3A+B)+3n(2A+B)]\frac{1}{5}\left[(-2)^n(3A+B) + 3^n(2A+B)\right]
  • (E)
    15[(−2)n(3A−B)+3n(2A−B)]\frac{1}{5}\left[(-2)^n(3A-B) + 3^n(2A-B)\right]

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