Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1997. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 24)

KALLOS, G., The generalization of Pascal's triangle from algebraic point of view

16 Gábor Kallós Proof. We get this result by induction immediately from the construction of the triangle. Lemma 2. Let us consider an expression n^aj

1 • • • am*-i > which i'o + «1 + • • • + im-2 + i

m-1 = n

• Then we can find this expression with some coefficient as a part of the element with the same weight in the n-th row of the general triangle. Proof. Let us assume indirectly that this expression does not exist in the n-th row of the general triangle as a part of the element with corresponding weight. We should get this expression from parts of elements of the previous row (

ni0-1 „»1 . . . 'm-2 ím-l „l'o-ti-l . . . tm-2 ««-1 V^O "1 m—2 "m —1 ' "0 "1 m — Z ™m — 1 1 a0 al m — 2 m —

1 ' 0 1 m —

2 m —

1 with multiplication (by ao, ai

,..., a

m_2

, a

m-i )• Thus, these parts of the elements can't exist in the previous row. Proceeding backwards with this method we conclude that in the first line some digits of the base-number do not exist, and this is a contradiction. Lemma 3. For the coefficient e of the expression ea

LQ a\

x • • -(L™l{ with io + ú + •"' + im-1 = n in the n-th row of the general triangle we have e = si io!il!'"tm-l! • Proof. We prove with induction. In the first row the statement is true. Let us now assume that in the n— 1-th row there axe the following expressions as parts of the elements: eoflS

0"

1«!

1---^, eiflSS

11-"«!::; e^a^-.-CrrN with coefficients (n — 1)! (n — 1)! e0 = T~- TTT

7-i : r 7 ei (»0 - l)!*i!---*m-i! ' i

0!(h - 1)! • •-im-i! «o!»l!- • -(im-i - 1)! (by the induction assumption). Thus, the coefficient e of the expression ea

lQ a

1^ • • • a™l[ is the sum of the coefficients e* (0 < i < m — 1) (n-l)! e = e

0 + ei + f- e

m_i = — («o-l)!(ti-l)!---(»m-i-l)!