Theorem.  Let the column-vector of functions of,n

                                                  T
               a(n):=, [a[0](n), a[1](n), a[2](n)] , be defined

                                                             T
                 by the initial conditions:, a(0) = [1, 0, 0]

          and a(n+1)=A(n)a(n), where A(n) is the, 3,  by , 3,  matrix

       3 n        3 n (n + 1)          3         7 n + 4     3 n
[[1/2, --- + 1/2, -----------], [- ---------, - ---------, - --- + 1/4],
        4              8           2 (n + 1)    4 (n + 1)     8

         3          3 n         5 n + 2
    [----------, ----------, - ---------]]
              2           2    8 (n + 1)
     2 (n + 1)   4 (n + 1)

                                                                    T
           In other words, a(n)=A(n-1)A(n-2)...A(0) times, [1, 0, 0]

                                      Let

                                  3                                 2
          F(n, k) = binomial(n, k)  (a[0](n) + a[1](n) k + a[2](n) k )

                   We have, for every non-negative integer n

                               -----
                                \
                                 )   F(n, k) = 1
                                /
                               -----
                                 k



                          Proof: We cleverly construct

                          /               3                 3
                        3 |      a[0](n) k         a[1](n) k  (6 n + 8 - 4 k)
G(n, k) = binomial(n, k)  |-1/2 ------------ + 1/8 --------------------------
                          |                3                         3
                          \     (n + 1 - k)               (n + 1 - k)

                    3     2                               2 \
           a[2](n) k  (3 n  - 3 n - 6 + 6 n k + 10 k - 4 k )|
     + 1/8 -------------------------------------------------|
                                        3                   |
                             (n + 1 - k)                    /

                              with the motive that

                 F(n + 1, k) - F(n, k) = G(n, k + 1) - G(n, k)

                                    (check!)

          and the identity follows upon summing with respect to k. QED