chapter  8
2A Simple Example
Pages 7

Remark 8.5. For m = −1 we can apply (2.3) (with c replaced by c− d) to get

( n

k

) (−1)n−k 1

c + kd =

dn[−1]n [c + nd, d]n+1

= dn(−1)n[n]n [c + nd, d]n+1

or 1

c(c + d) · · · (c + nd) = 1

dn[n]n

(−1)k(nk) c + kd

,

i.e., the partial fraction for a polynomial with equidistant roots. For d = 1 we even get

1 (c)n+1

= 1

[c + n]n+1 =

1 [n]n

(−1)k(nk) c + k

, (8.11)

i.e., the partial fraction for a polynomial with consecutive integral roots.