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.
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