1(1)
{ 0 , 1 , 16 , 81 , . . . , n 4 , . . . } \{0,1,16,81,...,n^4,...\} { 0,1,16,81,...,n4,...}
设 G { k ( k + 1 ) ( k + 2 ) ( k + 3 ) } = A ( x ) G\{k(k+1)(k+2)(k+3)\}=A(x) G{
k(k+1)(k+2)(k+3)}=A(x),则
∫ 0 x t 2 A ( t ) d x = ∑ k = 1 ∞ ∫ 0 x k ( k + 1 ) ( k + 2 ) ( k + 3 ) t k + 2 d t = ∑ k = 1 ∞ k ( k + 1 ) ( k + 2 ) x k + 3 = x 3 ∑ k = 1 ∞ k ( k + 1 ) ( k + 2 ) x k = x 3 G { k ( k + 1 ) ( k + 2 ) } = 6 x 4 ( 1 − x ) 4 \int^x_0t^2A(t)dx =\sum_{k=1}^{\infty}\int_0^x k(k+1)(k+2)(k+3)t^{k+2}dt\\=\sum_{k=1}^{\infty}k(k+1)(k+2)x^{k+3}\\=x^3\sum_{k=1}^{\infty}k(k+1)(k+2)x^k\\=x^3G\{k(k+1)(k+2)\}\\=\frac{6x^4}{(1-x)^4} ∫0xt2A(t)dx=k=1∑∞∫0xk(k+1)(k+2)(k+3)tk+2dt=k=1∑∞k(k+1)(k+2)xk+3=x3k=1∑∞k(k+1)(k+2)xk=x3G{
k(k+1)(k+2)}=(1−x)46x4
所以 x 2 A ( x ) = ( 6 x 4 ( 1 − x ) 4 ) ′ = 24 x 3 ( 1 − x ) 5 x^2A(x)=(\frac{6x^4}{(1-x)^4})'=\frac{24x^3}{(1-x)^5} x2A(x)=((1−x)46x4)′=(1−x)524x3
A ( x ) = 24 x ( 1 − x ) 5 A(x)=\frac{24x}{(1-x)^5} A(x)=(1−x)524x
k ( k + 1 ) ( k + 2 ) ( k + 3 ) = k 4 + 6 k 3 + 11 k 2 + 6 k k(k+1)(k+2)(k+3)=k^4+6k^3+11k^2+6k k(k+1)(k+2)(k+3)=k4+6k3+11k2+6k
G { k 4 } = G { k ( k + 1 ) ( k + 2 ) ( k + 3 ) } − 6 G { k 3 } − 11 G { k 2 } − 6 G { k } = 24 x ( 1 − x ) 5 − 6 x 3 + 4 x 2 + x ( 1 − x ) 4 − 11 x ( 1 + x ) ( 1 − x ) 3 − 6 x ( 1 − x ) 2 = x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 G\{k^4\}=G\{k(k+1)(k+2)(k+3)\}-6G\{k^3\}-11G\{k^2\}-6G\{k\}\\ =\frac{24x}{(1-x)^5}-6\frac{x^3+4x^2+x}{(1-x)^4}-11\frac{x(1+x)}{(1-x)^3}-6\frac{x}{(1-x)^2}\\ =\frac{x^4 + 11 x^3 + 11 x^2 + x}{(1-x)^5} G{
k4}=G{
k(k+1)(k+2)(k+3)}−6G{
k3}−11G{
k2}−6G{
k}=(1−x)524x−6(1−x)4x3+4x2+x−11(1−x)3x(1+x)−6(1−x)2x=(1−x)5x4+11x3+11x2+x
1(2)
{ ( 3 3 ) ( 4 3 ) , . . . , ( n + 3 3 ) } \{\binom33\binom43,...,\binom{n+3}{3}\} {
(33)(34),...,(3n+3)}
即 { ( 3 0 ) ( 4 1 ) , . . . , ( n + 3 n ) } \{\binom30\binom41,...,\binom{n+3}{n}\} {
(03)(14),...,(nn+3)}
k = 3 , G { ( n + 3 n ) } = 1 ( 1 − x ) 4 k=3,G\{\binom{n+3}{n}\}=\frac{1}{(1-x)^4} k=3,G{
(nn+3)}=(1−x)41
1(3)
{ 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , . . . } \{1,0,2,0,3,0,4,0,...\} { 1,0,2,0,3,0,4,0,...}
A ( x ) = 1 + 0 x 1 + 2 x 2 + 0 x 3 + 3 x 4 + 0 x 4 + 4 x 6 . . . = 1 + 2 x 2 + 3 x 4 + 4 x 6 . . . ( k + 1 ) x 2 k . . . A(x)=1+0x^1+2x^2+0x^3+3x^4+0x^4+4x^6...\\=1+2x^2+3x^4+4x^6...(k+1)x^{2k}... A(x)=1+0x1+2x2+0x3+3x4+0x4+4x6...=1+2x2+3x4+4x6...(k+1)x2k...
利用高中生方法，错位相减，
x 2 A ( x ) = x 2 + 2 x 4 + 3 x 6 . . . ( k + 1 ) x 2 k + 2 . . . ( 1 − x 2 ) A ( x ) = 1 + x 2 + x 4 + x 6 . . . ( 1 − x 2 ) A ( x ) = 1 1 − x 2 A ( x ) = ( 1 1 − x 2 ) 2 x^2A(x)=x^2+2x^4+3x^6...(k+1)x^{2k+2}...\\ (1-x^2)A(x)=1+x^2+x^4+x^6...\\ (1-x^2)A(x)=\frac{1}{1-x^2}\\ A(x)=(\frac{1}{1-x^2})^2 x2A(x)=x2+2x4+3x6...(k+1)x2k+2...(1−x2)A(x)=1+x2+x4+x6...(1−x2)A(x)=1−x21A(x)=(1−x21)2
1(4)
{ 1 , k , k 2 , k 3 , . . . , } \{1,k,k^2,k^3,...,\} { 1,k,k2,k3,...,}
A ( x ) = 1 + k x + k 2 x 2 . . . = 1 1 − k x A(x)=1+kx+k^2x^2...=\frac{1}{1-kx} A(x)=1+kx+k2x2...=1−kx1
2(1)
1 4 + 2 4 + . . . + n 4 1^4+2^4+...+n^4 14+24+...+n4
根据1(1)题，
G { k 4 } = x 4 + 11 x 3 + 11 x 2 + x ( 1 − x ) 5 G\{k^4\}=\frac{x^4 + 11 x^3 + 11 x^2 + x}{(1-x)^5} G{
k4}=(1−x)5x4+11x3+11x2+x
此时 a k = k 4 a_k=k^4 ak=k4,令 b n = 1 4 + 2 4 + . . . + n 4 = ∑ k = 0 n a k b_n=1^4+2^4+...+n^4=\sum_{k=0}^na_k bn=14+24+...+n4=∑k=0nak,则 { b n } \{b_n\} {
bn}生成函数为
B ( x ) = A ( x ) 1 − x = ( x + 11 x 2 + 11 x 3 + x 4 ) ( 1 − x ) 6 = ( x + 11 x 2 + 11 x 3 + x 4 ) ∑ n = 1 ∞ ( n + 5 n ) x n B(x)=\frac{A(x)}{1-x}\\ =\frac{(x+11x^2+11x^3+x^4)}{(1-x)^6}\\ =(x+11x^2+11x^3+x^4)\sum_{n=1}^\infty \binom{n+5}{n}x^n B(x)=1−xA(x)=(1−x)6(x+11x2+11x3+x4)=(x+11x2+11x3+x4)n=1∑∞(nn+5)xn
比较两边 x n x^n xn的系数，
1 4 + 2 4 + . . . + n 4 = ( n − 1 + 5 n − 1 ) + 11 ( n − 2 + 5 n − 2 ) + 11 ( n − 3 + 5 n − 3 ) + ( n − 4 + 5 n − 4 ) = ( n + 4 ) ( n + 3 ) ( n + 2 ) ( n + 1 ) n 120 + 11 ( n + 3 ) ( n + 2 ) ( n + 1 ) n ( n − 1 ) 120 + 11 ( n + 2 ) ( n + 1 ) n ( n − 1 ) ( n − 2 ) 120 + ( n + 1 ) n ( n − 1 ) ( n − 2 ) ( n − 3 ) 120 = 1 30 n ( n + 1 ) ( 2 n + 1 ) ( 6 n 3 + 9 n 2 + n − 1 ) 1^4+2^4+...+n^4\\= \binom{n-1+5}{n-1}+11\binom{n-2+5}{n-2}+11\binom{n-3+5}{n-3}+\binom{n-4+5}{n-4}\\ =\frac{(n+4)(n+3)(n+2)(n+1)n}{120}+11\frac{(n+3)(n+2)(n+1)n(n-1)}{120}+11\frac{(n+2)(n+1)n(n-1)(n-2)}{120}+\frac{(n+1)n(n-1)(n-2)(n-3)}{120}\\ =\frac{1}{30}n(n+1)(2n+1)(6n^3+9n^2+n-1) 14+24+...+n4=(n−1n−1+5)+11(n−2n−2+5)+11(n−3n−3+5)+(n−4n−4+5)=120(n+4)(n+3)(n+2)(n+1)n+11120(n+3)(n+2)(n+1)n(n−1)+11120(n+2)(n+1)n(n−1)(n−2)+120(n+1)n(n−1)(n−2)(n−3)=301n(n+1)(2n+1)(6n3+9n2+n−1)
2(2)
1 × 2 + 2 × 3 + . . . + n × ( n + 1 ) 1\times 2+2\times 3+...+n\times (n+1) 1×2+2×3+...+n×(n+1)
{ n ( n + 1 ) } \{n(n+1)\} {
n(n+1)}的生成函数为
A ( x ) = ∑ k = 0 n a k x k A(x)=\sum_{k=0}^na_kx^k A(x)=k=0∑nakxk
其中 a k = k ( k + 1 ) a_k=k(k+1) ak=k(k+1)
则
∑ k = 0 n a k x k = 2 x ( 1 − x ) 3 \sum_{k=0}^na_kx^k=\frac{2x}{(1-x)^3} k=0∑nakxk=(1−x)32x
令 b n = 1 × 2 + 2 × 3... n × ( n + 1 ) b_n=1\times 2+2\times 3...n\times(n+1) bn=1×2+2×3...n×(n+1),则
b n = ∑ k = 0 n a k b_n=\sum_{k=0}^na_k bn=k=0∑nak
{ b n } \{b_n\} {
bn}生成函数
B ( x ) = A ( x ) 1 − x = 2 x ( 1 − x ) 4 = 2 x ∑ i = 0 n ( i + 3 i ) x i B(x)=\frac{A(x)}{1-x}=\frac{2x}{(1-x)^4}\\ =2x\sum_{i=0}^n\binom{i+3}{i}x^i B(x)=1−xA(x)=(1−x)42x=2xi=0∑n(ii+3)xi
比较两边 x n x^n xn的系数
1 × 2 + 2 × 3... n × ( n + 1 ) = b n = 2 ( n + 2 n − 1 ) = n ( n + 1 ) ( n + 2 ) 3 1\times 2+2\times 3...n\times(n+1)=b_n=2\binom{n+2}{n-1}=\frac{n(n+1)(n+2)}{3} 1×2+2×3...n×(n+1)=bn=2(n−1n+2)=3n(n+1)(n+2)
4
设序列 { a n } \{a_n\} {
an}的生成函数为 4 − 3 x ( 1 − x ) ( 1 + x − x 3 ) \frac{4-3x}{(1-x)(1+x-x^3)} (1−x)(1+x−x3)4−3x,但 b 0 = a 0 , b 1 = a 1 − a 0 , . . . , b n = a n − a n − 1 b_0=a_0,b_1=a_1-a_0,...,b_n=a_n-a_{n-1} b0=a0,b1=a1−a0,...,bn=an−an−1，求序列 { b n } \{b_n\} {
bn}的生成函数。
∑ k = 0 n b k = a n \sum_{k=0}^nb_k=a_n k=0∑nbk=an
A ( x ) = G { a n } = 4 − 3 x ( 1 − x ) ( 1 + x − x 3 ) = G { b n } 1 − x A(x)=G\{a_n\}=\frac{4-3x}{(1-x)(1+x-x^3)}=\frac{G\{b_n\}}{1-x} A(x)=G{ an}=(1−x)(1+x−x3)4−3x=1−xG{ bn}
G { b n } = 4 − 3 x 1 + x − x 3 G\{b_n\}=\frac{4-3x}{1+x-x^3} G{ bn}=1+x−x34−3x
5
已知生成函数 3 − 9 x 1 − x − 56 x 2 \frac{3-9x}{1-x-56x^2} 1−x−56x23−9x,求对应的序列 { a n } \{a_n\} {
an}
3 − 9 x 1 − x − 56 x 2 = 3 − 9 x ( 8 x − 1 ) ( − 7 x − 1 ) = a 8 x − 1 + b ( − 7 x − 1 ) \frac{3-9x}{1-x-56x^2}\\=\frac{3-9x}{(8x-1)(-7x-1)}\\=\frac{a}{8x-1}+\frac{b}{(-7x-1)} 1−x−56x23−9x=(8x−1)(−7x−1)3−9x=8x−1a+(−7x−1)b
解得
= 1 1 − 8 x + − 2 − 7 x − 1 =\frac{1}{1-8x}+\frac{-2}{-7x-1} =1−8x1+−7x−1−2
a n = 1 × 8 n − 2 × ( − 7 ) n a_n=1\times 8^n-2\times (-7)^n an=1×8n−2×(−7)n
6.
有红,黄,蓝,白球各两个,绿,紫,黑球各3个,从中取出10个球,试问有多少种不同的取法？
M 红 = M 黄 = M 蓝 = M 白 = { 0 , 1 , 2 } , M 绿 = M 紫 = M 黑 = { 0 , 1 , 2 , 3 } M_{红}=M_{黄}=M_{蓝}=M_{白}=\{0,1,2\},M_{绿}=M_{紫}=M_{黑}=\{0,1,2,3\} M红=M黄=M蓝=M白={
0,1,2},M绿=M紫=M黑={
0,1,2,3}
( 1 + x + x 2 ) 4 ( 1 + x + x 2 + x 3 ) 3 (1+x+x^2)^4(1+x+x^2+x^3)^3 (1+x+x2)4(1+x+x2+x3)3
x 10 x^{10} x10的系数：678
怎么算出678？
( 1 + x + x 2 ) 4 ( 1 + x + x 2 + x 3 ) 3 = ( 1 − x 3 1 − x ) 4 ( 1 − x 4 1 − x ) 3 = ( 1 − x 3 ) 4 ( 1 − x 4 ) 3 ( 1 − x ) 7 = ( 1 − x 3 ) 4 ( 1 − x 4 ) 3 ∑ i = 0 n ( 7 + i − 1 i ) x i (1+x+x^2)^4(1+x+x^2+x^3)^3 \\=(\frac{1-x^3}{1-x})^4(\frac{1-x^4}{1-x})^3 \\=\frac{(1-x^3)^4(1-x^4)^3}{(1-x)^7} \\=(1-x^3)^4(1-x^4)^3\sum_{i=0}^n\binom{7+i-1}{i}x^i (1+x+x2)4(1+x+x2+x3)3=(1−x1−x3)4(1−x1−x4)3=(1−x)7(1−x3)4(1−x4)3=(1−x3)4(1−x4)3i=0∑n(i7+i−1)xi
组合成 x 10 x^{10} x10
( 7 + 10 − 1 10 ) − ( 4 1 ) ( 7 + 7 − 1 7 ) + ( 4 2 ) ( 7 + 4 − 1 4 ) − ( 4 3 ) ( 7 + 1 − 1 1 ) − ( 3 1 ) ( 7 + 6 − 1 6 ) + ( 3 2 ) ( 7 + 2 − 1 2 ) + ( 4 1 ) ( 3 1 ) ( 7 + 3 − 1 3 ) − ( 4 2 ) ( 3 1 ) ( 7 + 0 − 1 0 ) = 678 \binom{7+10-1}{10}-\binom{4}{1}\binom{7+7-1}{7}+\binom{4}{2}\binom{7+4-1}{4}-\binom{4}{3}\binom{7+1-1}{1}\\-\binom{3}{1}\binom{7+6-1}{6}+\binom{3}{2}\binom{7+2-1}{2} \\+\binom{4}{1}\binom{3}{1}\binom{7+3-1}{3}-\binom{4}{2}\binom{3}{1}\binom{7+0-1}{0} \\=678 (107+10−1)−(14)(77+7−1)+(24)(47+4−1)−(34)(17+1−1)−(13)(67+6−1)+(23)(27+2−1)+(14)(13)(37+3−1)−(24)(13)(07+0−1)=678
7.
口袋中有白球5个，红球3个，黑球2个，每次从中取5个，有多少种取法？
记 M 白 = { 0 , 1 , 2 , 3 , 4 , 5 } M_{白}=\{0,1,2,3,4,5\} M白={
0,1,2,3,4,5}, M 红 = { 0 , 1 , 2 , 3 } M_{红}=\{0,1,2,3\} M红={
0,1,2,3}, M 黑 = { 0 , 1 , 2 } M_{黑}=\{0,1,2\} M黑={
0,1,2}
( 1 + x + x 2 + x 3 + x 4 + x 5 ) ( 1 + x + x 2 + x 3 ) ( 1 + x + x 2 ) = ( 1 + x + x 2 + x 3 + x 4 + x 5 ) ( x 5 + 2 x 4 + 3 x 3 + 3 x 2 + 2 x + 1 ) (1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3)(1+x+x^2) \\=(1+x+x^2+x^3+x^4+x^5)(x^5+2x^4+3x^3+3x^2+2x+1) (1+x+x2+x3+x4+x5)(1+x+x2+x3)(1+x+x2)=(1+x+x2+x3+x4+x5)(x5+2x4+3x3+3x2+2x+1)
x 5 x^5 x5项系数： 1 × 1 + 2 × 1 + 3 × 1 + 3 × 1 + 2 × 1 + 1 × 1 = 12 1\times 1+2\times 1+3\times 1+3\times 1+2\times 1+1\times 1=12 1×1+2×1+3×1+3×1+2×1+1×1=12
8.
求1,3,5,7,9这5个数字组成的 n n n位数个数，要求其中3和7出现的次数为偶数，其他数字出现的次数无限制
记 M 1 = M 5 = M 9 = { 0 , 1 , 2 , . . . , n } M_1=M_5=M_9=\{0,1,2,...,n\} M1=M5=M9={
0,1,2,...,n}, M 3 = M 7 = { 0 , 2 , 4 , . . . , } M_3=M_7=\{0,2,4,...,\} M3=M7={
0,2,4,...,}
生成函数：
( 1 + x 2 2 ! + x 4 4 ! . . . ) 2 ( 1 + x + x 2 2 ! . . . ) 3 = e ( 3 x ) ( e ( x ) + e ( − x ) 2 ) 2 = e ( 5 x ) 4 + e ( 3 x ) 2 + e ( x ) 4 = 1 4 ∑ n = 0 ∞ ( 5 n + 2 × 3 n + 1 ) x n n ! (1+\frac{x^2}{2!}+\frac{x^4}{4!}...)^2(1+x+\frac{x^2}{2!}...)^3 \\=e(3x)(\frac{e(x)+e(-x)}{2})^2 \\=\frac{e(5x)}{4}+\frac{e(3x)}{2}+\frac{e(x)}{4} \\=\frac 14\sum_{n=0}^\infty(5^n+2\times 3^n+1)\frac{x^n}{n!} (1+2!x2+4!x4...)2(1+x+2!x2...)3=e(3x)(2e(x)+e(−x))2=4e(5x)+2e(3x)+4e(x)=41n=0∑∞(5n+2×3n+1)n!xn
其中， x n n ! \frac{x^n}{n!} n!xn系数为
a n = 1 4 ( 5 n + 2 × 3 n + 1 ) a_n=\frac 14(5^n+2\times 3^n+1) an=41(5n+2×3n+1)
9.
用3个1,2个2,5个3这10个数字能构成多少个偶的4位数？
最后一位必定是2，原问题可以转化为：用3个1、1个2、5个3这9个数能构成多少3位数？
M 1 = { 0 , 1 , 2 , 3 } , M 2 = { 0 , 1 } , M 3 = { 0 , 1 , 2 , 3 , 4 , 5 } M_1=\{0,1,2,3\},M_2=\{0,1\},M_3=\{0,1,2,3,4,5\} M1={
0,1,2,3},M2={
0,1},M3={
0,1,2,3,4,5}
( 1 + x + x 2 2 ! + x 3 3 ! ) ( 1 + x ) ( 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + x 5 5 ! ) (1+x+\frac {x^2}{2!}+\frac{x^3}{3!})(1+x)(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}) (1+x+2!x2+3!x3)(1+x)(1+x+2!x2+3!x3+4!x4+5!x5)
x 3 x^3 x3项系数为 20 6 \frac{20}{6} 620，则 x 3 3 ! \frac{x^3}{3!} 3!x3项系数为20,
20个偶四位数
12
把正整数8写成3个非负整数之和，要求 n 1 ≤ 3 , n 2 ≤ 3 , n 3 ≤ 6 n_1\le 3,n_2\le 3,n_3\le 6 n1≤3,n2≤3,n3≤6,有多少种不同方案？
M n 1 = M n 2 = { 0 , 1 , 2 , 3 } , M n 3 = { 0 , 1 , 2 , 3 , 4 , 5 , 6 } M_{n_1}=M_{n_2}=\{0,1,2,3\},M_{n_3}=\{0,1,2,3,4,5,6\} Mn1=Mn2={
0,1,2,3},Mn3={
0,1,2,3,4,5,6}
( 1 + x + x 2 + x 3 ) 2 ( 1 + x + x 2 + x 3 + x 4 + x 5 + x 6 ) (1+x+x^2+x^3)^2(1+x+x^2+x^3+x^4+x^5+x^6) (1+x+x2+x3)2(1+x+x2+x3+x4+x5+x6)
( 1 + 2 x + 3 x 2 + 4 x 3 + 3 x 4 + 2 x 5 + x 6 ) ( 1 + x + x 2 + x 3 + x 4 + x 5 + x 6 ) (1+2x+3x^2+4x^3+3x^4+2x^5+x^6)(1+x+x^2+x^3+x^4+x^5+x^6) (1+2x+3x2+4x3+3x4+2x5+x6)(1+x+x2+x3+x4+x5+x6)
x 8 x^8 x8的系数为：
3 × 1 + 4 × 1 + 3 × 1 + 2 × 1 + 1 × 1 = 13 3\times 1+4\times 1+3\times 1+2\times 1+1\times 1=13 3×1+4×1+3×1+2×1+1×1=13
18
设有砝码重为1g的3个，重为2g的4个，重为4g的2个，则能称出多少种重量？各有多少种方案？
M 1 = { 0 , 1 , 2 , 3 } , M 2 = { 0 , 1 , 2 , 3 , 4 } , M 4 = { 0 , 1 , 2 } M_1=\{0,1,2,3\},M_2=\{0,1,2,3,4\},M_4=\{0,1,2\} M1={ 0,1,2,3},M2={ 0,1,2,3,4},M4={ 0,1,2}
生成函数为
( 1 + x + x 2 + x 3 ) ( 1 + x 2 + x 4 + x 6 + x 8 ) ( 1 + x 4 + x 8 ) = 1 + x + 2 x 2 + 2 x 3 + 3 x 4 + 3 x 5 + 4 x 6 + 4 x 7 + 5 x 8 + 5 x 9 + 5 x 10 + 5 x 11 + 4 x 12 + 4 x 13 + 3 x 14 + 3 x 15 + 2 x 16 + 2 x 17 + x 18 + x 19 (1+x+x^2+x^3)(1+x^2+x^4+x^6+x^8)(1+x^4+x^8) \\=1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8+5x^9+5x^{10}+5x^{11}+4x^{12}+4x^{13}+3x^{14}+3x^{15}+2x^{16}+2x^{17}+x^{18}+x^{19} (1+x+x2+x3)(1+x2+x4+x6+x8)(1+x4+x8)=1+x+2x2+2x3+3x4+3x5+4x6+4x7+5x8+5x9+5x10+5x11+4x12+4x13+3x14+3x15+2x16+2x17+x18+x19
能称出20种重量。
19
求 x 1 + 2 x 2 + 4 x 3 = 21 x_1+2x_2+4x_3=21 x1+2x2+4x3=21的正整数解个数。
x 1 x_1 x1为奇数， M 1 = { 1 , 3 , 5 , . . . , } M_1=\{1,3,5,...,\} M1={ 1,3,5,...,}
M 2 = { 2 , 4 , . . . , } , M 3 = { 4 , 8 , 12 , . . . , } M_2=\{2,4,...,\},M_3=\{4,8,12,...,\} M2={ 2,4,...,},M3={ 4,8,12,...,}
生成函数为
( x + x 3 + x 5 . . . ) ( x 2 + x 4 + x 6 . . . ) ( x 4 + x 8 . . . ) = x 7 ( 1 + x 2 + x 4 . . . ) 2 ( 1 + x 4 + x 8 . . . ) = x 7 1 ( 1 − x 2 ) 2 1 1 − x 4 = x 7 1 ( 1 − x 2 ) 2 ( 1 + x 2 ) 2 ( 1 − x 2 ) 2 ( 1 − x 4 ) 3 = x 7 ( 1 + x 2 ) 2 ( 1 − x 4 ) 3 = ( x 11 + 2 x 9 + x 7 ) 1 ( 1 − x 4 ) 3 = ( x 11 + 2 x 9 + x 7 ) ∑ k = 0 ∞ ( k + 2 2 ) x 4 k (x+x^3+x^5...)(x^2+x^4+x^6...)(x^4+x^8...) \\=x^7(1+x^2+x^4...)^2(1+x^4+x^8...) \\=x^7\frac{1}{(1-x^2)^2}\frac{1}{1-x^4} \\=x^7\frac{1}{(1-x^2)^2}\frac{(1+x^2)^2(1-x^2)^2}{(1-x^4)^3} \\=x^7\frac{(1+x^2)^2}{(1-x^4)^3} \\=(x^{11}+2x^9+x^7)\frac{1}{(1-x^4)^3} \\=(x^{11}+2x^9+x^7)\sum_{k=0}^\infty \binom{k+2}{2}x^{4k} (x+x3+x5...)(x2+x4+x6...)(x4+x8...)=x7(1+x2+x4...)2(1+x4+x8...)=x7(1−x2)211−x41=x7(1−x2)21(1−x4)3(1+x2)2(1−x2)2=x7(1−x4)3(1+x2)2=(x11+2x9+x7)(1−x4)31=(x11+2x9+x7)k=0∑∞(2k+2)x4k
x 21 x^{21} x21的系数: 2 x 9 ( 5 2 ) x 12 = 20 x 21 2x^9\binom{5}{2}x^{12}=20x^{21} 2x9(25)x12=20x21
正整数解个数20
P66(3) 用生成函数方法做
n n n位四进制数中，2和3出现偶数次的序列的数目记为 f ( n ) f(n) f(n),求 f ( n ) f(n) f(n)满足的递推关系
M 2 = { 0 , 2 , 4 , . . . , } , M 3 = { 0 , 2 , 4 , . . . , } , M 0 = { 0 , 1 , 2 , . . . , } , M 1 = { 0 , 1 , 2 , . . . , } M_2=\{0,2,4,...,\},M_3=\{0,2,4,...,\},M_0=\{0,1,2,...,\},M_1=\{0,1,2,...,\} M2={
0,2,4,...,},M3={
0,2,4,...,},M0={
0,1,2,...,},M1={
0,1,2,...,}
生成函数
( 1 + x 2 2 ! + x 4 4 ! . . . ) 2 ( 1 + x + x 2 2 ! . . . ) 2 = ( e ( x ) + e ( − x ) 2 ) 2 e ( 2 x ) = e ( 2 x ) e ( 2 x ) + 2 + e ( − 2 x ) 4 = e ( 4 x ) + 2 e ( 2 x ) + 1 4 (1+\frac{x^2}{2!}+\frac{x^4}{4!}...)^2(1+x+\frac{x^2}{2!}...)^2 \\=(\frac{e(x)+e(-x)}{2})^2e(2x) \\=e(2x)\frac{e(2x)+2+e(-2x)}{4} \\=\frac{e(4x)+2e(2x)+1}{4} (1+2!x2+4!x4...)2(1+x+2!x2...)2=(2e(x)+e(−x))2e(2x)=e(2x)4e(2x)+2+e(−2x)=4e(4x)+2e(2x)+1
其中， x n n ! \frac{x^n}{n!} n!xn项系数 a n = 1 4 × 4 n + 1 2 × 2 n a_n=\frac{1}{4}\times 4^n+\frac{1}{2}\times 2^n an=41×4n+21×2n
p83 简单变式
从 { n ⋅ a , n ⋅ b , n ⋅ c } \{n\cdot a,n\cdot b,n\cdot c\} { n⋅a,n⋅b,n⋅c}中取出 n n n个字母，要求 a a a的个数为奇数，有多少种取法？
M a = { 1 , 3 , 5 , . . . } M_a=\{1,3,5,...\} Ma={
1,3,5,...}, M b = M c = { 0 , 1 , 2 , . . . , } M_b=M_c=\{0,1,2,...,\} Mb=Mc={
0,1,2,...,}
( x + x 3 + x 5 . . . ) ( 1 + x + x 2 . . . ) 2 = x ( 1 + x 2 + x 4 . . . ) ( 1 + x + x 2 . . . ) 2 = x 1 1 − x 2 1 ( 1 − x ) 2 = x 2 ( 1 1 + x + 1 1 − x ) ( 1 ( 1 − x ) 2 ) = x 2 ( 1 ( 1 + x ) ( 1 − x ) 2 + 1 ( 1 − x ) 3 ) = . . . = x 8 ( 1 1 + x + 1 1 − x + 2 ( 1 − x ) 2 + 4 ( 1 − x ) 3 ) = x 8 ∑ i = 0 ∞ ( ( − 1 ) i + 1 + 2 ( i + 1 i ) + 4 ( i + 2 i ) ) x i (x+x^3+x^5...)(1+x+x^2...)^2 \\=x(1+x^2+x^4...)(1+x+x^2...)^2 \\=x\frac{1}{1-x^2}\frac{1}{(1-x)^2} \\=\frac x2(\frac 1{1+x}+\frac 1{1-x})(\frac 1{(1-x)^2}) \\=\frac x2(\frac 1{(1+x)(1-x)^2}+\frac 1{(1-x)^3}) \\=... \\=\frac x8 (\frac 1{1+x}+\frac 1{1-x}+\frac{2}{(1-x)^2}+\frac{4}{(1-x)^3}) \\=\frac x8\sum_{i=0}^\infty ((-1)^i+1+2\binom{i+1}i+4\binom{i+2}{i})x^i (x+x3+x5...)(1+x+x2...)2=x(1+x2+x4...)(1+x+x2...)2=x1−x21(1−x)21=2x(1+x1+1−x1)((1−x)21)=2x((1+x)(1−x)21+(1−x)31)=...=8x(1+x1+1−x1+(1−x)22+(1−x)34)=8xi=0∑∞((−1)i+1+2(ii+1)+4(ii+2))xi
x n x^n xn系数为
1 8 [ ( − 1 ) n − 1 + 1 + 2 ( i − 1 + 1 i − 1 ) + 4 ( i − 1 + 2 i − 1 ) ] = 1 8 [ ( − 1 ) n − 1 + 1 + 2 n + 2 n ( n + 1 ) ] \frac 18[(-1)^{n-1}+1+2\binom{i-1+1}{i-1}+4\binom{i-1+2}{i-1}] \\=\frac 18[(-1)^{n-1}+1+2n+2n(n+1)] 81[(−1)n−1+1+2(i−1i−1+1)+4(i−1i−1+2)]=81[(−1)n−1+1+2n+2n(n+1)]
P31 6
安排5人去3个学校参观，每个学校至少1人，共有多少种安排方案？
M 1 = M 2 = M 3 = { 1 , 2 , 3 , . . . , } M_1=M_2=M_3=\{1,2,3,...,\} M1=M2=M3={ 1,2,3,...,}
生成函数为
( x 1 ! + x 2 2 ! + . . . ) 3 = ( e ( x ) − 1 ) 3 = e ( 3 x ) − 3 e ( 2 x ) + 3 e ( x ) − 1 (\frac{x}{1!}+\frac{x^2}{2!}+...)^3=(e(x)-1)^3=e(3x)-3e(2x)+3e(x)-1 (1!x+2!x2+...)3=(e(x)−1)3=e(3x)−3e(2x)+3e(x)−1
x 5 5 ! \frac{x^5}{5!} 5!x5系数： 3 5 − 3 × 2 5 + 3 = 150 3^5-3\times 2^5+3=150 35−3×25+3=150
P31 8
有7种小球，每个小球内有1~7个星星。一次活动中，主办方随机发放礼品盒，每个盒里放两个这样的小球，那么共有多少种这样的礼品盒？
M 1 = M 2 = M 3 = M 4 = M 5 = M 6 = M 7 = { 0 , 1 , 2 , . . . , ∞ } M_1=M_2=M_3=M_4=M_5=M_6=M_7=\{0,1,2,...,\infty\} M1=M2=M3=M4=M5=M6=M7={ 0,1,2,...,∞}
( 1 + x + x 2 . . . ) 7 = ( 1 1 − x ) 7 = G { ( n + 6 n ) } (1+x+x^2...)^7=(\frac{1}{1-x})^7=G\{\binom{n+6}{n}\} (1+x+x2...)7=(1−x1)7=G{ (nn+6)}
n = 2 , ( 8 2 ) = 28 n=2,\binom{8}{2}=28 n=2,(28)=28