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5.1 Determinants Which Represent Particular Polynomials  175

          But

                                   j − 1     j − 1    (j−i+1)
                   a + a i−1,j =         +           u

                    ij             i − 1     i − 2

                                     j − 1     j − 1    (j−i+2)
                                −          +           v
                                     i − 2     i − 3

                                   j     (j−i+1)     j     (j−i+2)
                              =         u      −         v
                                  i − 1            i − 2
                              = a i,j+1 .                           (5.1.13)
          Hence,
                    C + C = C j+1 ,                                 (5.1.14)

                          ∗
                     j    j
                               n


                         A =       C 1 C 2 ··· C C j+1 ··· C n−1 C n ,



                          n                 j
                                                              n
                              j=1
                       (n+1)
                     A      = − C 1 C 2 ··· C j C j+1 ··· C n−1 C n+1 .  (5.1.15)


                       n+1,n
                                                              n
          Hence,
                                n
                        (n+1)

                 A + A       =
                   n    n+1,n     C 1 C 2 ··· (C − C j+1 ) C j+1 ··· C n
                                              j
                                                                  n
                               j=1
                                  n


                                               ∗
                             = −
                                    C 1 C 2 ··· C ··· C n
                                               j
                                 j=1
                             =0
          by Theorem 3.1 on cyclic dislocations and generalizations in Section 3.1,
          which proves (a).
            Expanding A n+1 by the two elements in its last row,
                                                (n+1)

                         A n+1 =(u − nv )A n − vA
                                                n+1,n

                              =(u − nv )A n + vA
                                                n
                                         u   nv


                              = v A +      −      A n ,
                                         v    v
                                    n

                        yA n+1   y         y    nv

                              =     A +      −
                        v  n+1  v n   n    y    v   A n
                                yA     1  y  2
                              =    n  +
                                 v  n   v  n  A n

                              = D   yA n
                                     v n

                              = D 2  yA n−1
                                     v n−1

                                     yA n−r+1
                              = D r           ,  0 ≤ r ≤ n
                                     v n−r+1
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