111 新北市高中聯招
版主: thepiano
111 新北市高中聯招
請參考附件
- 附加檔案
-
- 111 新北市高中聯招.pdf
- (538.98 KiB) 已下載 346 次
-
- 111 新北市高中聯招_答案.pdf
- (312.23 KiB) 已下載 345 次
Re: 111 新北市高中聯招
填充第 3 題
可以這樣拆
1 / (1^3 + 2^3 + ... + k^3)
= 4 / [k^2 * (k + 1)^2]
= 4[(2k + 3) / (k + 1)^2 - (2k - 1) / k^2]
可以這樣拆
1 / (1^3 + 2^3 + ... + k^3)
= 4 / [k^2 * (k + 1)^2]
= 4[(2k + 3) / (k + 1)^2 - (2k - 1) / k^2]
Re: 111 新北市高中聯招
第 4 題
g(x) = f(7x + 8) - (7x)^3
g(1) = f(15) - 7^3
g(2) = f(22) - 8 * 7^3
g(3) = f(29) - 27 * 7^3
g(4) = f(36) - 64 * 7^3
由巴貝奇定理
g(4) - 3g(3) + 3g(2) - g(1) = 0
f(36) - 64 * 7^3 - 3f(29) + 81 * 7^3 + 3f(22) - 24 * 7^3 - f(15) + 7^3 = 0
f(36) = 6 * 7^3 + 3f(29) - 3f(22) + f(15) = 2058 - 93 + 69 - 15 = 2019
g(x) = f(7x + 8) - (7x)^3
g(1) = f(15) - 7^3
g(2) = f(22) - 8 * 7^3
g(3) = f(29) - 27 * 7^3
g(4) = f(36) - 64 * 7^3
由巴貝奇定理
g(4) - 3g(3) + 3g(2) - g(1) = 0
f(36) - 64 * 7^3 - 3f(29) + 81 * 7^3 + 3f(22) - 24 * 7^3 - f(15) + 7^3 = 0
f(36) = 6 * 7^3 + 3f(29) - 3f(22) + f(15) = 2058 - 93 + 69 - 15 = 2019
Re: 111 新北市高中聯招
計算第 1 題
f(x) = Q_1(x)(x - 1)^2 + ax + b
= Q_2(x)(x + 1)^2 + bx + a
= Q_3(x)(x - 1)^2(x + 1)^2 + cx^3 + dx^2 + ex
ac ≠ 0
f(1) = a + b = c + d + e
f(-1) = a - b = - c + d - e
可解出 a = d,b = c + e
f '(1) = a = 3c + 2d + e
f '(-1) = b = 3c - 2d + e
d = 3c + 2d + e
c + e = 3c - 2d + e
可解出 d = c,e = -4c
R(x) = cx^3 + dx^2 + ex = cx^3 + cx^2 - 4cx = 0
x(x^2 + x - 4) = 0
x = 0 or (-1 ± √17)/2
f(x) = Q_1(x)(x - 1)^2 + ax + b
= Q_2(x)(x + 1)^2 + bx + a
= Q_3(x)(x - 1)^2(x + 1)^2 + cx^3 + dx^2 + ex
ac ≠ 0
f(1) = a + b = c + d + e
f(-1) = a - b = - c + d - e
可解出 a = d,b = c + e
f '(1) = a = 3c + 2d + e
f '(-1) = b = 3c - 2d + e
d = 3c + 2d + e
c + e = 3c - 2d + e
可解出 d = c,e = -4c
R(x) = cx^3 + dx^2 + ex = cx^3 + cx^2 - 4cx = 0
x(x^2 + x - 4) = 0
x = 0 or (-1 ± √17)/2
Re: 111 新北市高中聯招
計算第 2 題
(1) 當 n + 1 為完全平方數,且有 m 個正因數,易知 m 是正奇數
[√(n + 1)] = [√n] + 1
Σ[(n + 1) / k] (k = 1 ~ n + 1) = Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) = [√n] + Σ[n / k] (k = 1 ~ n) + (m + 1)
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) 和 [√n] + Σ[n / k] (k = 1 ~ n) 同奇或同偶
(2) 當 n + 1 不為完全平方數,且有 m 個正因數,易知 m 是正偶數
[√(n + 1)] = [√n]
Σ[(n + 1) / k] (k = 1 ~ n + 1) = Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) = [√n] + Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) 和 [√n] + Σ[n / k] (k = 1 ~ n) 同奇或同偶
(3) 而 n = 1 時,[√1] + [1 / 1] = 2,是偶數
故對任意正整數 n, [√n] + Σ[n / k] (k = 1 ~ n) 必為偶數
(1) 當 n + 1 為完全平方數,且有 m 個正因數,易知 m 是正奇數
[√(n + 1)] = [√n] + 1
Σ[(n + 1) / k] (k = 1 ~ n + 1) = Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) = [√n] + Σ[n / k] (k = 1 ~ n) + (m + 1)
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) 和 [√n] + Σ[n / k] (k = 1 ~ n) 同奇或同偶
(2) 當 n + 1 不為完全平方數,且有 m 個正因數,易知 m 是正偶數
[√(n + 1)] = [√n]
Σ[(n + 1) / k] (k = 1 ~ n + 1) = Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) = [√n] + Σ[n / k] (k = 1 ~ n) + m
[√(n + 1)] + Σ[(n + 1) / k] (k = 1 ~ n + 1) 和 [√n] + Σ[n / k] (k = 1 ~ n) 同奇或同偶
(3) 而 n = 1 時,[√1] + [1 / 1] = 2,是偶數
故對任意正整數 n, [√n] + Σ[n / k] (k = 1 ~ n) 必為偶數