APPENDI X

Appendix

0.2(1):

Set A = [a][a - 1] • • • [a - n + 2] for n 1 and A = 1 for n = 1. We have

[n]V"

+ «'

a—n-f-1

a

n - 1

) = A(^-n[a -

n

+ 1] + va~n+l[n])

To get (1) we have to show that the term in the last parentheses is equal to [a + 1].

Well, we have more generally

v-b[a]+va[b]

ya—b _ y—a—b _ yd+b I yd—b

- 7 , ~ 1

V — V

[a + 6]

for all integers a and 6.

0.2(3):

For r = 1 the left hand side is equal to

= 1 - 1 = 0.

We get for r 1 using 0.2(1)

r r -i r

i = 0

i=0

r - 1

r - 1

+ v

r-i+1

r - 1

i - 1

i= 0

r - 1

2

r - 1

j ' = o

r - 1

The first sum is equal to 0 by induction, the second one is equal to

r - 1

^ 2 r - l ^

(

_

1 ) V

( ( r - l ) - - l )

i=o

r - 1

3

hence also 0 by induction.