组合数学引论ch3答案
时间:2025-02-23
时间:2025-02-23
|ÜêÆÚØ1nÙ‰Y
5·35·( 2)13= C5·35·213,2.x5y13 Xê´µC1818
8·38·( 2)10=C5·38·210.x8y10 Xê´µC1818
nnnn1n+1n+1n+13(2). ª >=[(n+1)++++0123
nn+1···+]n n+1n+1n+1n+11=[++···++]12nn+11(2n+1 1)==m>8. m
=
m(m 1)(m 2)
,3
m m1∴a2
=
m(m 1)
,=m,
+b
m
+c
m
=a
m(m 1)(m 2)
m3
21
3+b am2+2a 3b+6cm,=am+b
m(m 1)
+cm
1
b a=0,2a 3b+6c=0,∴a=1,∴a=6,b=6,c=1,13+23+···+n3
=[(6+632
nnn(6+6+)]
321
12n12n=6(++···+)+6(++···+)+
333222 12n
···+(++···+)
111 n+1n+1n+1=6+6+
432 n+2n+1=6+
42
=
n2(n+1)2 1 1
1222+)+(6+6+)+···+
1321
10.(1)y²µ Ý n iÎG¥§0Ñyóêg iÎG ê f(n) §ÑyÛêg iÎG ê g(n) §@o§k
f(n)=g(n 1)+2f(n 1)=3n 1 f(n 1)+2f(n 1)=3n 1+f(n 1),
n 1 1)+2=3n+1(3f(n)=3n 1+3n 2+···+31+f(1)=3(2) ª > ±n) Ý Ý n iÎG¥§0Ñyóê
g iÎG êoÚ§=Ó1(1)¯"
15. >=
n 1
+
n 1
n 3
kk 1k
n 2n 2n 1n 3=++
kk 1k 1k n 3n 3n 2n 1n 3=+++
kk 1k 1k 1k=m>16. ª===
m1
m2m3
s=
t
=
r,s,t≥0r+s+t=n
r,s,t≥0r+s+t=n
n s t
m3m1+m2
n t
t
m1+m2+m3
n
nnkn
18.y²µ∵(1+x)=k=0x,
k ªü>Ó ¦ , µ
nnk 1n 1
=k=0kn(1+x)x,
k
nnkn 1
x,nx(1+x)=k=0kkü>2g¦ §Ó x=1µ >=n(n+1)2n 2,
m>=,
n
k=1k
2nk
∴y..