题目

Problem

In this question use the standard results for summations.

(a) Show that for all positive integers $n$

\begin{align*} &\,\sum_{r=1}^n (12r^2+2r-3)\\ =&\,An^3+Bn^2 \end{align*}

where $A$ and $B$ are integers to be determined.

(4)

(b) Hence determine the value of $n$ for which

\begin{align*} &\,\sum_{r=1}^{2n}r^3\\ &\,-\sum_{r=1}^n(12r^2+2r-3)\\ =&\,270 \end{align*}

(4)

题目中文翻译

本题使用标准求和公式。

(a) 证明对所有正整数 $n$ ，均有

\begin{align*} &\,\sum_{r=1}^n (12r^2+2r-3)\\ =&\,An^3+Bn^2 \end{align*}

其中 $A$ 和 $B$ 为待求的整数。

(4)

(b) 由此确定满足以下等式的 $n$ 值：

\begin{align*} &\,\sum_{r=1}^{2n}r^3\\ &\,-\sum_{r=1}^n(12r^2+2r-3)\\ =&\,270 \end{align*}

(4)

解答

展开

把原求和拆成平方和、自然数和与常数项之和，分别代入标准公式，再展开合并。一次项会恰好抵消，留下题目要求的三次项与二次项。

展开

Using the standard results for summations,

\begin{align*} &\,\sum_{r=1}^n(12r^2+2r-3)\\ =&\,12\sum_{r=1}^n r^2\\ &\,\hspace{2pt}+2\sum_{r=1}^n r\\ &\,\hspace{4pt}-3\sum_{r=1}^n1\\ =&\,12\big(\frac{n(n+1)(2n+1)}6\big)\\ &\,\hspace{2pt}+2\big(\frac{n(n+1)}2\big)\\ &\,\hspace{4pt}-3n\\ =&\,2n(n+1)(2n+1)\\ &\,\hspace{2pt}+n(n+1)-3n\\ =&\,4n^3+6n^2+2n\\ &\,\hspace{2pt}+n^2+n-3n\\ =&\,4n^3+7n^2. \end{align*}

Comparing this with $An^3+Bn^2$ gives

\boxed{A=4,\qquad B=7}.

展开

承接 (a)，用 $4n^3+7n^2$ 代替第二个求和；第一个求和则使用立方和公式，并注意上限是 $2n$ 。所得方程只含 $n^4$ 与 $n^2$ ，把它看作关于 $n^2$ 的二次方程求解，再用 $n$ 为正整数筛选答案。

展开

From part (a),

\begin{align*} &\,\sum_{r=1}^n(12r^2+2r-3)\\ =&\,4n^3+7n^2. \end{align*}

Also,

\begin{align*} \sum_{r=1}^{2n}r^3 =&\,\big(\frac{2n(2n+1)}2\big)^2\\ =&\,n^2(2n+1)^2\\ =&\,4n^4+4n^3+n^2. \end{align*}

Therefore,

\begin{align*} &\,\sum_{r=1}^{2n}r^3\\ &\,\hspace{2pt}-\sum_{r=1}^n(12r^2+2r-3)\\ =&\,270. \end{align*}

Substituting the expressions above gives

\begin{align*} &\,4n^4+4n^3+n^2\\ &\,\hspace{2pt}-(4n^3+7n^2)=270\\ &\,4n^4-6n^2=270\\ &\,2n^4-3n^2=135\\ &\,2n^4-3n^2-135=0\\ &\,(2n^2+15)(n^2-9)=0. \end{align*}

Since $n$ is a positive integer, $2n^2+15\neq0$ and $n^2=9$ gives

\boxed{n=3}.