遇到这种该怎么做呢

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遇到这种该怎么做呢
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遇到这种该怎么做呢
遇到这种该怎么做呢
 

遇到这种该怎么做呢

  1. a(n+2) = 5/3 * a(n+1) - 2/3 * a(n)

    a(n+2) - a(n+1) = 2/3 * a(n+1) - 2/3 * a(n)

    b(n+1) = 2/3 * b(n)

    所以bn是等比数列, q = 2/3

  2. b(1) = a(2) - a(1) = 3

    b(n...

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    1. a(n+2) = 5/3 * a(n+1) - 2/3 * a(n)

      a(n+2) - a(n+1) = 2/3 * a(n+1) - 2/3 * a(n)

      b(n+1) = 2/3 * b(n)

      所以bn是等比数列, q = 2/3

    2. b(1) = a(2) - a(1) = 3

      b(n) = b(1) * q^(n-1) = 3 * (2/3)^(n-1)

      b(1) + b(2) +...+ b(n-1) = a(n) - a(1)

      Sb(n-1) = a(n) - 1

      Sb(n-1) = b(1) * (1-q^(n-1))/(1-q) = 3 * (1-(2/3)^(n-1))/(1-2/3)

      Sb(n-1) = 9 * ( 1 - (2/3)^(n-1) ) = a(n) - 1

      a(n) = 9 - 9 * (2/3)^(n-1) + 1

      a(n) = 10 - 9*(2/3)^(n-1)

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