答案速览
题号
答案
1
3 3 3
2
恒等式成立，第二项应为 y m x j y^mx^j y m x j
3
不等式成立，实际上严格大于 1 1 1
4
α = ( a − a 2 − 4 a ) / 2 , β = ( a + a 2 − 4 a ) / 2 \alpha=(a-\sqrt{a^2-4a})/2,\ \beta=(a+\sqrt{a^2-4a})/2 α = ( a − a 2 − 4 a ) /2 , β = ( a + a 2 − 4 a ) /2
5
唯一解 f ( x ) = 3 x f(x)=3x f ( x ) = 3 x
6
f ( x ) 2 + g ( x ) 2 = 1 f(x)^2+g(x)^2=1 f ( x ) 2 + g ( x ) 2 = 1
逐题解析
第 1 题
题目
已知 p p p 3 3 3
∏ k = 1 p ( 1 + 2 cos 2 k π p ) \prod_{k=1}^{p}\left(1+2\cos\frac{2k\pi}{p}\right) k = 1 ∏ p ( 1 + 2 cos p 2 k π ) 的值。
解答
令 ζ = e 2 π i / p \zeta=e^{2\pi i/p} ζ = e 2 π i / p k = p k=p k = p 3 3 3
1 + ζ k + ζ − k = ζ − k 1 − ζ 3 k 1 − ζ k . 1+\zeta^k+\zeta^{-k}
=\zeta^{-k}\frac{1-\zeta^{3k}}{1-\zeta^k}. 1 + ζ k + ζ − k = ζ − k 1 − ζ k 1 − ζ 3 k . 因为 p > 3 p>3 p > 3 3 k m o d p 3k\bmod p 3 k mod p 1 ≤ k ≤ p − 1 1\le k\le p-1 1 ≤ k ≤ p − 1 1 1 1 ζ − ∑ k = ζ − p ( p − 1 ) / 2 = 1 \zeta^{-\sum k}=\zeta^{-p(p-1)/2}=1 ζ − ∑ k = ζ − p ( p − 1 ) /2 = 1 3 3 3
第 2 题
题目
已知 m , n m,n m , n x > 0 x>0 x > 0 y > 0 y>0 y > 0 x + y = 1 x+y=1 x + y = 1
∑ i = 0 m − 1 ( n − 1 + i i ) x n y i + ∑ j = 0 n − 1 ( m − 1 + j j ) y m x j = 1. \sum_{i=0}^{m-1}\binom{n-1+i}{i}x^ny^i+
\sum_{j=0}^{n-1}\binom{m-1+j}{j}y^mx^j=1. i = 0 ∑ m − 1 ( i n − 1 + i ) x n y i + j = 0 ∑ n − 1 ( j m − 1 + j ) y m x j = 1. 解答
来源正文及概率解释均要求第二和式为 ∑ j = 0 n − 1 ( m − 1 + j j ) y m x j \sum_{j=0}^{n-1}\binom{m-1+j}{j}y^mx^j ∑ j = 0 n − 1 ( j m − 1 + j ) y m x j
独立投掷正面概率 x x x y y y n n n m m m m + n − 1 m+n-1 m + n − 1 n n n i i i 0 ≤ i < m 0\le i<m 0 ≤ i < m ( n − 1 + i i ) x n y i \binom{n-1+i}{i}x^ny^i ( i n − 1 + i ) x n y i m m m ( m − 1 + j j ) y m x j \binom{m-1+j}{j}y^mx^j ( j m − 1 + j ) y m x j 1 1 1
第 3 题
题目
已知 n ∈ N + n\in\mathbb N^+ n ∈ N + n ≥ 2 n\ge2 n ≥ 2 x i > 0 x_i>0 x i > 0 i = 1 , 2 , … , n i=1,2,\ldots,n i = 1 , 2 , … , n
x 1 x 2 + x 2 x 3 + ⋯ + x n − 1 x n + x n x 1 ≥ 1. x_1^{x_2}+x_2^{x_3}+\cdots+x_{n-1}^{x_n}+x_n^{x_1}\ge1. x 1 x 2 + x 2 x 3 + ⋯ + x n − 1 x n + x n x 1 ≥ 1. 解答
若某个 x i ≥ 1 x_i\ge1 x i ≥ 1 1 1 1 0 < x i < 1 0<x_i<1 0 < x i < 1 x n + 1 = x 1 x_{n+1}=x_1 x n + 1 = x 1
对 u > 0 , 0 < t < 1 u>0,0<t<1 u > 0 , 0 < t < 1 s t s^t s t ( 1 + u ) t ≤ 1 + t u (1+u)^t\le1+tu ( 1 + u ) t ≤ 1 + t u u = ( 1 − x i ) / x i , t = x i + 1 u=(1-x_i)/x_i,t=x_{i+1} u = ( 1 − x i ) / x i , t = x i + 1
x i x i + 1 ≥ x i x i + x i + 1 − x i x i + 1 > x i x i + x i + 1 ≥ x i x 1 + ⋯ + x n . x_i^{x_{i+1}}\ge
\frac{x_i}{x_i+x_{i+1}-x_ix_{i+1}}
>\frac{x_i}{x_i+x_{i+1}}
\ge\frac{x_i}{x_1+\cdots+x_n}. x i x i + 1 ≥ x i + x i + 1 − x i x i + 1 x i > x i + x i + 1 x i ≥ x 1 + ⋯ + x n x i . 对所有 i i i 1 1 1
第 4 题
题目
已知数列 { x n } \{x_n\} { x n } { y n } \{y_n\} { y n } x 1 = 1 x_1=1 x 1 = 1 x n + y n = a n − 1 x_n+y_n=an-1 x n + y n = an − 1 a > 4 a>4 a > 4 a ∈ N + a\in\mathbb N^+ a ∈ N + x n + 1 x_{n+1} x n + 1 x 1 , x 2 , … , x n , y 1 , y 2 , … , y n x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n x 1 , x 2 , … , x n , y 1 , y 2 , … , y n α , β \alpha,\beta α , β
x n = ⌊ α n ⌋ , y n = ⌊ β n ⌋ . x_n=\lfloor\alpha n\rfloor,\qquad y_n=\lfloor\beta n\rfloor. x n = ⌊ α n ⌋ , y n = ⌊ β n ⌋ . 解答
取表中的 α < β \alpha<\beta α < β α + β = a , α β = a \alpha+\beta=a,\alpha\beta=a α + β = a , α β = a 1 / α + 1 / β = 1 1/\alpha+1/\beta=1 1/ α + 1/ β = 1 a > 4 a>4 a > 4 a 2 − 4 a a^2-4a a 2 − 4 a ( a − 3 ) 2 (a-3)^2 ( a − 3 ) 2 ( a − 2 ) 2 (a-2)^2 ( a − 2 ) 2 1 < α < 2 < β 1<\alpha<2<\beta 1 < α < 2 < β
先证明两个取整序列互补。对任意整数 N ≥ 0 N\ge0 N ≥ 0 N N N
⌊ N + 1 α ⌋ + ⌊ N + 1 β ⌋ = N . \left\lfloor\frac{N+1}{\alpha}\right\rfloor+
\left\lfloor\frac{N+1}{\beta}\right\rfloor=N. ⌊ α N + 1 ⌋ + ⌊ β N + 1 ⌋ = N . 因为括号内两个数都是非整数，和恰为 N + 1 N+1 N + 1 N N N N − 1 N-1 N − 1
⌊ α n ⌋ + ⌊ β n ⌋ = a n − 1. \lfloor\alpha n\rfloor+\lfloor\beta n\rfloor=an-1. ⌊ α n ⌋ + ⌊ β n ⌋ = an − 1. ⌊ α ⌋ = 1 \lfloor\alpha\rfloor=1 ⌊ α ⌋ = 1 ⌊ β n ⌋ > ⌊ α n ⌋ \lfloor\beta n\rfloor>\lfloor\alpha n\rfloor ⌊ β n ⌋ > ⌊ α n ⌋ n n n ⌊ α ( n + 1 ) ⌋ \lfloor\alpha(n+1)\rfloor ⌊ α ( n + 1 )⌋ n n n ⌊ α ( n + 1 ) ⌋ \lfloor\alpha(n+1)\rfloor ⌊ α ( n + 1 )⌋
第 5 题
题目
求所有函数 f : ( 0 , + ∞ ) → ( 0 , + ∞ ) f:(0,+\infty)\to(0,+\infty) f : ( 0 , + ∞ ) → ( 0 , + ∞ ) x x x
f ( f ( x ) ) = 12 x − f ( x ) . f(f(x))=12x-f(x). f ( f ( x )) = 12 x − f ( x ) . 解答
固定 x > 0 x>0 x > 0 u 0 = x , u n + 1 = f ( u n ) u_0=x,u_{n+1}=f(u_n) u 0 = x , u n + 1 = f ( u n ) u n > 0 u_n>0 u n > 0 u n + 2 + u n + 1 = 12 u n u_{n+2}+u_{n+1}=12u_n u n + 2 + u n + 1 = 12 u n 3 , − 4 3,-4 3 , − 4 u n = A 3 n + B ( − 4 ) n u_n=A3^n+B(-4)^n u n = A 3 n + B ( − 4 ) n B ≠ 0 B\ne0 B = 0 4 > 3 4>3 4 > 3 u n < 0 u_n<0 u n < 0 B = 0 B=0 B = 0 u 1 = 3 u 0 u_1=3u_0 u 1 = 3 u 0 f ( x ) = 3 x f(x)=3x f ( x ) = 3 x
第 6 题
题目
已知在 R \mathbb R R f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x ) x , y x,y x , y
f ( x + y ) = f ( x ) f ( y ) − g ( x ) g ( y ) , g ( x + y ) = f ( x ) g ( y ) + g ( x ) f ( y ) , f(x+y)=f(x)f(y)-g(x)g(y),\qquad
g(x+y)=f(x)g(y)+g(x)f(y), f ( x + y ) = f ( x ) f ( y ) − g ( x ) g ( y ) , g ( x + y ) = f ( x ) g ( y ) + g ( x ) f ( y ) , 且 f ′ ( 0 ) = 0 f'(0)=0 f ′ ( 0 ) = 0 f 2 ( x ) + g 2 ( x ) = 1 f^2(x)+g^2(x)=1 f 2 ( x ) + g 2 ( x ) = 1
解答
先令 y = 0 y=0 y = 0
( f ( 0 ) − 1 ) f ( x ) − g ( 0 ) g ( x ) = 0 , g ( 0 ) f ( x ) + ( f ( 0 ) − 1 ) g ( x ) = 0. (f(0)-1)f(x)-g(0)g(x)=0,\quad
g(0)f(x)+(f(0)-1)g(x)=0. ( f ( 0 ) − 1 ) f ( x ) − g ( 0 ) g ( x ) = 0 , g ( 0 ) f ( x ) + ( f ( 0 ) − 1 ) g ( x ) = 0. 存在 x x x ( f ( x ) , g ( x ) ) ≠ ( 0 , 0 ) (f(x),g(x))\ne(0,0) ( f ( x ) , g ( x )) = ( 0 , 0 ) ( f ( 0 ) − 1 ) 2 + g ( 0 ) 2 = 0 (f(0)-1)^2+g(0)^2=0 ( f ( 0 ) − 1 ) 2 + g ( 0 ) 2 = 0 f ( 0 ) = 1 , g ( 0 ) = 0 f(0)=1,g(0)=0 f ( 0 ) = 1 , g ( 0 ) = 0
对 y y y y = 0 y=0 y = 0 c = g ′ ( 0 ) c=g'(0) c = g ′ ( 0 ) f ′ ( 0 ) = 0 f'(0)=0 f ′ ( 0 ) = 0 f ′ ( x ) = − c g ( x ) , g ′ ( x ) = c f ( x ) f'(x)=-cg(x),g'(x)=cf(x) f ′ ( x ) = − c g ( x ) , g ′ ( x ) = c f ( x ) ( f 2 + g 2 ) ′ = 0 (f^2+g^2)'=0 ( f 2 + g 2 ) ′ = 0 1 1 1