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Solutions to Problems; Chapter 1 261

From the recursion relation,

(Y + zPn(x, Y, z + 1) - (x + Y)P*(X + 1, Y, 4 + (2 - z>Pn(x, Y, z)

= (y+z)[(a:+z+l)(y+%+l)P n - 1 ( x,Y,%+2)-(%+1)2pn-l(a:,Y,%+1)1

• (x + Y>[(X + z + l)(Y + z)Pn-1(x + Ly, z + 1) - z2Prl-1(x + 1, y, %)I
  
  
  • (x - z)[(x + %)(Y + z)Pn.-1(x, y, z + 1) - z2pn-l(X,Y, 41.
    Using Q+i(x, y, z + 1) = Qn-i(t, y, z) = 0, we have that
  
  
  
  

(i) (Y + z + l)P,-^1 ( x,y,z + 2) = (x + y)P,-1(x + l,y,z + 1) -

(x - % - l)Pn-1(x, y, z + 1)

(ii) (x--z)Pn-l(x, Y, %) = (x+y)P,-1(x+1, Y, z)--(~+z)P~-l(x, Y, z+l).

Substituting these into Qn(x, y, z) yields a linear combination of

P,-~(a: + 1, y, z + l), P,-1(x, y, z + 1) and Pn-l(z + 1, y, r), all of whose

coefficients vanish. Hence Qn(x, y, z) = 0 as required.

9.15. Each term in the expansion is a product

xf’ xfaxfJ... x;”

where the nonnegative integers ui satisfy

a2 + a3 +... +u,sn-1
01 + a2 + 03 +.. f + 0, = 12.
Conversely, for any choice of nonnegative integers satisfying these condi-
tions, there is a term in the expansion with these exponents. (Each factor
of the expression contributes an xi to the product; for like terms, there is
no cancellation. To build a term of the required type, start with the xi of
highest index i and work from right to left through the product.)
Let f(n) be the number of terms. Clearly, f(1) = 1 and f(2) = 2. In
selecting the ui, consider the possibility that the first equality holds in the
rth line, so that we have

a, = 0

an-1 I 1

&I-r+2 +... + a,-1^5 r -^2 1

(4

G-r+1 +... + a,-1 = P (B)

h-r I 1

a,-,+1 + anvr 5 2

1

cc>

a1 + ** * + unwr = n - T. J