第3題
可以這樣拆
\(\begin{align}
  & \frac{1}{{{1}^{3}}+{{2}^{3}}+\cdots +{{k}^{3}}} \\
& =\frac{4}{{{k}^{2}}{{\left( k+1 \right)}^{2}}} \\
& =4\left[ \frac{2k+3}{{{\left( k+1 \right)}^{2}}}-\frac{2k-1}{{{k}^{2}}} \right] \\
\end{align}\)

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

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


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

請寫一下您的完整做法