Concrete mathematics : a foundation for computer science

5.60 (‘c) z 4n/&K is the case m = n of
(my) z /gq(l + ;)n(l + G)?
A ANSWERS TO EXERCISES 519
5.61 Let [n/p] = q and n mod p = r. The polynomial identity (x + 1 )P -
xp + 1 (mod p) implies that
(x+ 1) pq+r i= (~+l)~(x~ +l)q (mod p).
The coefficient of x”’ on the left is (E). On the right it’s tk (,I,,) (z), which
is just ( m mbd ,) ( ,m~tpJ) because 0 6 r 3.)
5.63 This is S, = ~~=,(-4)k(~+~) = ~~=,(-4)nPk(2n~k). The denominator
of (5.74) is zero when z = -l/4, so we can’t simply plug into that formula.
The recurrence S, =I -2&-l -.SnP2 leads to the solution S, = (-l)n(2n+l).
5.64 ~,,,((;k) + (2;+,))/@+ 1) = &O (;$,)/(k+ 11, which is
A.& (g;:) = '",';;"
,
5.65 Multiply both sides by nn-’ and replace k by n - 1 - k to get
nk(n - k)! = (n - l)! Z(nkf’/k! - nk/(k - l.)!)
x (7Y)
n-1
k k=O
= (n-l)!nn/(n-l)!.
(The partial sums can, in fact, be found by Gosper’s algorithm.) Alternatively,
(2 knnPlekk! can be interpreted as the number of mappings of {l , . . . , n} into
itself with f (1)) . . . , f(k)distinctbutf(k+l) � {f(l),...,f(k)};summingonk
must give nn.
5.66 This is a “wa.lk the garden path” problem where there’s only one “obvious”
way to proceed at every step. First replace k - j by 1, then replace
[A] by k, getting
j&o (j:‘k) (A) y ’
I ,