题目

Problem

(a) Use the standard results for summations to show that

\begin{align*} &\,\sum_{r=1}^{n}(2r-1)^2\\ =&\,\frac n3(An^2+B) \end{align*}

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

(5)

(b) Use the answer to part (a) to determine the value of

\begin{align*} &\,25^2+27^2+29^2\\ &\,\hspace{2pt}+\cdots+95^2 \end{align*}

You must make your method clear.

(3)

题目中文翻译

(a) 使用求和标准公式证明

\begin{align*} &\,\sum_{r=1}^{n}(2r-1)^2\\ =&\,\frac n3(An^2+B) \end{align*}

其中 $A$ 和 $B$ 是待确定的整数。

(5)

(b) 利用 (a) 的答案确定

\begin{align*} &\,25^2+27^2+29^2\\ &\,\hspace{2pt}+\cdots+95^2 \end{align*}

的值。必须清楚说明方法。

(3)

解答

展开

先展开 $(2r-1)^2$ ，把原式拆成平方项之和、一次项之和与常数项之和。代入标准求和公式并提取公因式 $\frac n3$ ，最后与题目指定形式比较。

展开

Using the standard results for summations,

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

Therefore, comparing with $\dfrac n3(An^2+B)$ ,

\boxed{A=4,\qquad B=-1}.

展开

通项中的奇数是 $2r-1$ 。解 $2r-1=25$ 与 $2r-1=95$ ，确定所求范围是第 13 项至第 48 项，再用“前 48 项之和减去前 12 项之和”。

展开

Since

25=2(13)-1

and

95=2(48)-1,

the required sum consists of the 13th to the 48th odd squares. Using part (a),

\begin{align*} &\,25^2+27^2+\cdots+95^2\\ =&\,\sum_{r=1}^{48}(2r-1)^2\\ &\,\hspace{2pt}-\sum_{r=1}^{12}(2r-1)^2\\ =&\,\frac{48}{3}\big[4(48)^2-1\big]\\ &\,-\frac{12}{3}\big[4(12)^2-1\big]\\ =&\,16(9215)-4(575)\\ =&\,147440-2300\\ =&\,\boxed{145140}. \end{align*}