(a) 用数学归纳法证明：对于 $n \in \mathbb{Z}^+$， $\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2$

(b) 使用标准求和公式证明 $\sum_{r=1}^{n} r(r + 1)(r - 1) = \frac{1}{4}n(n + A)(n + B)(n + C)$ 其中 $A$、$B$、$C$ 为待确定的常数。

(c) 确定使下式成立的 $n$ 值 $3\sum_{r=1}^{n} r(r + 1)(r - 1) = 17\sum_{r=n}^{2n} r^2$

先验证 $n=1$。归纳步骤中，在前 $k$ 项立方和后加上 $(k+1)^3$，再提取 $\frac14(k+1)^2$；括号内恰好是 $(k+2)^2$，从而得到 $n=k+1$ 时的目标形式。

For $n=1$,

and

Hence the result is true for $n=1$.

Assume that the result is true for $n=k$, where $k\in\mathbb Z^+$. Thus

Then

This is the required result for $n=k+1$. Since the result is true for $n=1$, and truth for $n=k$ implies truth for $n=k+1$, it follows by mathematical induction that

for all $n\in\mathbb Z^+$.

先展开 $r(r+1)(r-1)=r^3-r$，再使用 (a) 的立方和结果与标准的一次方求和公式。提取 $\frac14n(n+1)$ 后，剩余二次式可分解为 $(n-1)(n+2)$。

Since

using part (a) and the standard result for $\sum r$ gives

Therefore one valid assignment of the constants is

where the order of $A,B,C$ may be interchanged.

先把从 $n$ 到 $2n$ 的平方和写成“前 $2n$ 项平方和减去前 $n-1$ 项平方和”，并用标准公式化简。然后代入 (b) 的结果，约去正整数条件下非零的公共因子，解所得二次方程并排除非正整数根。

Using the standard result for the sum of squares,

Using part (b), the given equation becomes

Since $n$ is a positive integer, $n(n+1)\neq0$, so this factor may be cancelled. Therefore,

Thus $n=-1/9$ or $n=52$. Since $n\in\mathbb Z^+$,