chapter  8
6 Joseph M. Santmyer’s Problem
Solution.
Pages 3

Solution. We have nonzero terms only for 0 ≤ k ≤ n−m− 1 and 0 ≤ j ≤ m− 1, hence the sum is finite.

We write the sum as

S = n−m−1∑

[k + j]k [k]k

· [2n−m− 2k − j − 3]m−1−j [m− 1− j]m−1−j

= n−m−1∑

1 [k]k

[k + j]k [2n−m− 2k − 3]m−1[m− 1]j [2n−m− 2k − 3]j [m− 1]m−1

= 1

[m− 1]m−1 n−m−1∑

[2n−m− 2k − 3]m−1 [k]k

[k + j]k[m− 1]j [2n−m− 2k − 3]j .