chapter  11
Sums of Type II(5,5,1)
Pages 6

Substitution of this into the sum of (11.2) yields, after changing the order of summation,

( 2n j

) [2n− a− b− 1]j [a]2n−j [b]2n−j ([c]j [d]j)2

× 2n∑ k=j

( 2n− j k − j

) [2n− k]j [c− j]k−j [d− j]k−j [c− j]2n−j−k

× [d− j]2n−j−k(−1)k−j . Now, we have [2n − k]j = 0 for j > 2n − k, so we can end summation for k = 2n− j. The inner sum then starts

( 2n− 2j

2n− k − j )

[2n− j]j · · · .