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

                            2                             3       2
   3 (n + 1)   (n + 1) (10 n  + 10 n + 3)  n (n + 1) (20 n  + 30 n  + 15 n + 2)
[[-----------, --------------------------, ------------------------------------
  2 (4 n + 3)    4 (2 n + 1) (4 n + 3)       4 (1 + 4 n) (2 n + 1) (4 n + 3)

                           2                      4       3      2
             5         14 n  + 12 n + 3       20 n  + 20 n  + 2 n  - 4 n - 1
    ], [- -------, - ---------------------, - -------------------------------],
          4 n + 3    2 (2 n + 1) (4 n + 3)    2 (1 + 4 n) (2 n + 1) (4 n + 3)

                                                    3      2
             5                  5 n              2 n  - 7 n  - 5 n - 1
    [-----------------, -------------------, -----------------------------]]
     (n + 1) (4 n + 3)  2 (n + 1) (4 n + 3)  2 (1 + 4 n) (4 n + 3) (n + 1)

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

                                      Let

                                  4                                 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

                          /
                        4 |
G(n, k) = binomial(n, k)  |
                          |
                          \

                 4                                 3        2       2
        a[0](n) k  (24 n k - 10 - 70 n + 4 k - 80 n  - 140 n  + 32 n  k)
    1/4 ---------------------------------------------------------------- + 1/4
                                                            4
                   (1 + 4 n) (2 n + 1) (4 n + 3) (n + 1 - k)

             4                 2       3          2           2       2  2
    a[1](n) k  (15 - 18 k + 6 k  - 80 n  k - 180 n  k + 32 n k  + 32 n  k

                              2       4        3    /
     - 112 n k + 105 n + 220 n  + 40 n  + 170 n )  /  ((1 + 4 n) (2 n + 1)
                                                  /

                         4                 4       2  3       4        3
    (4 n + 3) (n + 1 - k) ) + 1/4 a[2](n) k  (-80 k  n  + 10 n  + 240 n  k

                    4                  2              2        2  2         3
     + 50 k + 40 k n  + 248 n k + 400 n  k - 20 - 42 k  - 220 n  k  + 40 n k

           3  2          2       3                2        3       5    /
     + 32 k  n  - 176 n k  + 12 k  - 110 n - 195 n  - 115 n  + 20 n )  /  (
                                                                      /

                                               \
                                             4 |
    (1 + 4 n) (2 n + 1) (4 n + 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