3.1 二项式定理
定理 3.1.1（二项式定理）
设 n n n 为一正整数，则对任意的 x x x 和 y y y，有
( x + y ) n = y n + ( n 1 ) x y n − 1 + ( n 2 ) x 2 y n − 2 + ⋯ + ( n n − 1 ) x n − 1 y + x n (x + y)^n = y^n + \binom{n}{1}x y^{n-1} + \binom{n}{2}x^2 y^{n-2} + \cdots + \binom{n}{n-1}x^{n-1} y + x^n (x+y)n=yn+(1n)xyn−1+(2n)x2yn−2+⋯+(n−1n)xn−1y+xn
或等价地表示为：
( x + y ) n = ∑ r = 0 n ( n r ) x r y n − r . (x + y)^n = \sum_{r=0}^n \binom{n}{r} x^r y^{n-r}. (x+y)n=r=0∑n(rn)xryn−r.
定理 3.1.2
对一切实数 α \alpha α 和 x x x（ ∣ x ∣ < 1 |x| < 1 ∣x∣<1），有
( 1 + x ) α = ∑ r = 0 ∞ ( α r ) x r , (1 + x)^{\alpha} = \sum_{r=0}^{\infty} \binom{\alpha}{r} x^r, (1+x)α=r=0∑∞(rα)xr,
其中
( α r ) = α ( α − 1 ) ⋯ ( α − r + 1 ) r ! . \binom{\alpha}{r} = \frac{\alpha(\alpha - 1)\cdots(\alpha - r + 1)}{r!}. (rα)=r!α(α−1)⋯(α−r+1).
(1) α = − n \alpha=-n α=−n
( 1 + x ) − n = ∑ r = 0 ∞ ( − 1 ) r ( n + r − 1 r ) x r . (1 + x)^{-n} = \sum_{r=0}^{\infty} (-1)^r \binom{n + r - 1}{r} x^r. (1+x)−n=r=0∑∞(−1)r(rn+r−1)xr.
(2) α = 1 2 \alpha=\frac{1}{2} α=21
( 1 + x ) 1 2 = ∑ r = 0 + ∞ ( − 1 ) r − 1 ⋅ 1 r ⋅ 2 2 r − 1 ⋅ ( 2 r − 2 r − 1 ) x r . (1 + x)^{\frac{1}{2}} = \sum_{r=0}^{+\infty} (-1)^{r-1} \cdot \frac{1}{r \cdot 2^{2r-1}} \cdot \binom{2r - 2}{r - 1}x^r. (1+x)21=r=0∑+∞(−1)r−1⋅r⋅22r−11⋅(r−12r−2)xr.
3.2二项式系数
当 n , r n, r n,r 均为非负整数，且 n ≥ r n \geq r n≥r 时， ( n r ) \binom{n}{r} (rn) 有一些基本的性质：
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对称关系
( n r ) = ( n n − r ) \binom{n}{r} = \binom{n}{n-r} (rn)=(n−rn) -
递推关系
( n r ) = ( n − 1 r ) + ( n − 1 r − 1 ) ( n ≥ r ≥ 1 ) \binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1} \quad (n \geq r \geq 1) (rn)=(rn−1)+(r−1n−1)(n≥r≥1)
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单峰性
当 n n n 为偶数时，有
( n 0 ) < ( n 1 ) < ⋯ < ( n n 2 ) , ( n n 2 ) > ⋯ > ( n n − 1 ) > ( n n ) \binom{n}{0} < \binom{n}{1} < \cdots < \binom{n}{\frac{n}{2}}, \binom{n}{\frac{n}{2}} > \cdots > \binom{n}{n-1} > \binom{n}{n} (0n)<(1n)<⋯<(2nn),(2nn)>⋯>(n−1n)>(nn)当 n n n 为奇数时，有
( n 0 ) < ( n 1 ) < ⋯ < ( n n − 1 2 ) = ( n n + 1 2 ) , ⋯ \binom{n}{0} < \binom{n}{1} < \cdots < \binom{n}{\frac{n-1}{2}} = \binom{n}{\frac{n+1}{2}}, \cdots (0n)<(1n)<⋯<(2n−1n)=(2n+1n),⋯
3.3组合恒等式
等式 1
( n 0 ) + ( n 1 ) + ⋯ + ( n n ) = 2 n . \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n. (0n)+(1n)+⋯+(nn)=2n.
证明： 在二项式定理中令 x = y = 1 x = y = 1 x=y=1 即可。
等式 2
( n 0 ) + ( n 2 ) + ( n 4 ) + ⋯ = ( n 1 ) + ( n 3 ) + ( n 5 ) + ⋯ . \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots. (0n)+(2n)+(4n)+⋯=(1n)+(3n)+(5n)+⋯.
证明： 在二项式定理中令 x = − 1 , y = 1 x = -1, y = 1 x=−1,y=1，得
∑ k = 0 n ( − 1 ) k ( n k ) = 0. \sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0. k=0∑n(−1)k(kn)=0.
整理一下即得。
等式 3
1 ⋅ ( n 1 ) + 2 ⋅ ( n 2 ) + ⋯ + n ⋅ ( n n ) = n ⋅ 2 n − 1 . 1 \cdot \binom{n}{1} + 2 \cdot \binom{n}{2} + \cdots + n \cdot \binom{n}{n} = n \cdot 2^{n-1}. 1⋅(1n)+2⋅(2n)+⋯+n⋅(nn)=n⋅2n−1.
证明： 对等式
( 1 + x ) n = ∑ i = 0 n ( n i ) x i (1 + x)^n = \sum_{i=0}^{n} \binom{n}{i} x^i (1+x)n=i=0∑n(in)xi
两边在 x = 1 x = 1 x=1 处求导数，得
( 1 + x ) n \] x = 1 ′ = n ( 1 + x ) n − 1 ∣ x = 1 = n 2 n − 1 , \\left\[(1 + x)\^n\\right\]'_{x=1} = n(1 + x)\^{n-1}\\big\|_{x=1} = n2\^{n-1}, \[(1+x)n\]x=1′=n(1+x)n−1 x=1=n2n−1, \[ ∑ i = 0 n ( n i ) x i \] x = 1 ′ = ∑ i = 1 n i ( n i ) x i − 1 ∣ x = 1 = ∑ i = 1 n i ( n i ) , \\left\[\\sum_{i=0}\^{n} \\binom{n}{i} x\^i\\right\]'_{x=1} = \\sum_{i=1}\^{n} i \\binom{n}{i} x\^{i-1}\\big\|_{x=1} = \\sum_{i=1}\^{n} i \\binom{n}{i}, \[i=0∑n(in)xi\]x=1′=i=1∑ni(in)xi−1 x=1=i=1∑ni(in), 从而 1 ⋅ ( n 1 ) + 2 ⋅ ( n 2 ) + ⋯ + n ⋅ ( n n ) = n ⋅ 2 n − 1 . 1 \\cdot \\binom{n}{1} + 2 \\cdot \\binom{n}{2} + \\cdots + n \\cdot \\binom{n}{n} = n \\cdot 2\^{n-1}. 1⋅(1n)+2⋅(2n)+⋯+n⋅(nn)=n⋅2n−1. #### 等式 4 ( 0 k ) + ( 1 k ) + ⋯ + ( n k ) = ( n + 1 k + 1 ) . \\binom{0}{k} + \\binom{1}{k} + \\cdots + \\binom{n}{k} = \\binom{n+1}{k+1}. (k0)+(k1)+⋯+(kn)=(k+1n+1). #### 等式 5 ∑ k = 0 n ( n k ) 2 = ( 2 n n ) \\sum_{k=0}\^{n} \\binom{n}{k}\^2 = \\binom{2n}{n} k=0∑n(kn)2=(n2n) #### 等式 6 ∑ i = 0 m ( m i ) ( n r + i ) = ( m + n m + r ) \\sum_{i=0}\^{m} \\binom{m}{i} \\binom{n}{r+i} = \\binom{m+n}{m+r} i=0∑m(im)(r+in)=(m+rm+n) ### 3.4多项式定理 #### 定理 3.4.1 设 n n n 为正整数，则 ( x 1 + x 2 + ⋯ + x t ) n = ∑ ( n n 1 n 2 ⋯ n t ) x 1 n 1 x 2 n 2 ⋯ x t n t , (x_1 + x_2 + \\cdots + x_t)\^n = \\sum \\binom{n}{n_1 n_2 \\cdots n_t} x_1\^{n_1} x_2\^{n_2} \\cdots x_t\^{n_t}, (x1+x2+⋯+xt)n=∑(n1n2⋯ntn)x1n1x2n2⋯xtnt, 其中 ( n n 1 n 2 ⋯ n t ) = n ! n 1 ! n 2 ! ⋯ n t ! \\binom{n}{n_1 n_2 \\cdots n_t} = \\frac{n!}{n_1! n_2! \\cdots n_t!} (n1n2⋯ntn)=n1!n2!⋯nt!n! #### 定理 3.4.2 给定正整数 n ; t ; n 1 , n 2 , ⋯ , n t n; t; n_1, n_2, \\cdots, n_t n;t;n1,n2,⋯,nt，且 n 1 + n 2 + ⋯ + n t = n n_1 + n_2 + \\cdots + n_t = n n1+n2+⋯+nt=n，那么 ( n n 1 n 2 ⋯ n t ) = ( n − 1 ( n 1 − 1 ) n 2 ⋯ n t ) + ( n − 1 n 1 ( n 2 − 1 ) ⋯ n t ) + ⋯ + ( n − 1 n 1 n 2 ⋯ ( n t − 1 ) ) \\binom{n}{n_1 n_2 \\cdots n_t} = \\binom{n-1}{(n_1-1) n_2 \\cdots n_t} + \\binom{n-1}{n_1 (n_2-1) \\cdots n_t} + \\cdots + \\binom{n-1}{n_1 n_2 \\cdots (n_t-1)} (n1n2⋯ntn)=((n1−1)n2⋯ntn−1)+(n1(n2−1)⋯ntn−1)+⋯+(n1n2⋯(nt−1)n−1) ### 例题