Concrete mathematics : a foundation for computer science

They expect us to
check this
on a sheet of
scratch paper.
in Problem 1. Replacing T by -k - 2 gives the desired expansion,
5% = & (“:“) (1)&y& (7) (-k;2)~1.
,
5.2 BASIC PRACTICE 185
Now the (k + l)-’ can be absorbed into (z), as planned. In fact, it could
also be absorbed into (-kj- 2)p1. Double absorption suggests that even more
cancellation might be possible behind the scenes. Yes-expanding everything
in our new summand into factorials and going back to binomial coefficients
gives a formula that we can sum on k:
~t-l)j(mn++;,+l) c (;;l++;;;) (-n; ')
sm = (mE-t)! j>.
m! n!
= (m+n+l)!
,I m + n + l j
xc- I.(
n+l+j n ’
j20
JO
The sum over all integers j is zero, by (5.24). Hence -S, is the sum for j < 0.
To evaluate -S, for j < 0, let’s replace j by -k - 1 and sum for k 3 0:
m! n!
~(-l)frn,+“k’l) (-k;l)
sm = (m+n+l)! k>O
I . . I
;lp,y-k(m+;+ ‘> (“n”- ‘>
= (m+mnn+l)! k