CalcGospel 國際數學圖譜
返回

CIE 9231 2022 November Paper 11 Q3

A Level / CIE / FP1

CIE 9231 2022 Nov Paper 11 Paper · Question 3

题目

Problem

(a) By considering (2r+1)3−(2r−1)3(2r+1)^3-(2r-1)^3, use the method of differences to prove that

∑r=1nr2=16n(n+1)(2n+1).\sum_{r=1}^{n}r^2=\frac16n(n+1)(2n+1).
(5)

Let Sn=12+3×22+32+3×42+52+3×62+…+(2+(−1)n)n2S_n=1^2+3\times2^2+3^2+3\times4^2+5^2+3\times6^2+\ldots+\big(2+(-1)^n\big)n^2.

(b) Show that S2n=13n(2n+1)(an+b)S_{2n}=\frac13n(2n+1)(an+b), where aa and bb are integers to be determined.

(3)

(c) State the value of lim⁡n→∞S2nn3\displaystyle\lim_{n\to\infty}\frac{S_{2n}}{n^3}.

(1)
题目中文翻译

(a) 通过考虑 (2r+1)3−(2r−1)3(2r+1)^3-(2r-1)^3，用裂项相消法证明

∑r=1nr2=16n(n+1)(2n+1).\sum_{r=1}^{n}r^2=\frac16n(n+1)(2n+1).
(5)

设 Sn=12+3×22+32+3×42+52+3×62+…+(2+(−1)n)n2S_n=1^2+3\times2^2+3^2+3\times4^2+5^2+3\times6^2+\ldots+\big(2+(-1)^n\big)n^2。

(b) 证明 S2n=13n(2n+1)(an+b)S_{2n}=\frac13n(2n+1)(an+b)，其中 aa 和 bb 是待求的整数。

(3)

(c) 写出 lim⁡n→∞S2nn3\displaystyle\lim_{n\to\infty}\frac{S_{2n}}{n^3} 的值。

(1)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

展开相邻奇数的立方差，对两边求和。左边相消后只剩首尾，右边包含平方和。

答题过程

展开

Expanding,

(2r+1)3−(2r−1)3= (8r3+12r2+6r+1) −(8r3−12r2+6r−1)= 24r2+2.\begin{align*} (2r+1)^3-(2r-1)^3 =&\,(8r^3+12r^2+6r+1)\\[4mm] &\,\hspace{2pt}-(8r^3-12r^2+6r-1)\\[4mm] =&\,24r^2+2. \end{align*}

Summing from r=1r=1 to r=nr=n, the left side telescopes:

 (33−13)+(53−33) +(73−53)+⋯ +((2n−3)3−(2n−5)3) +((2n−1)3−(2n−3)3) +((2n+1)3−(2n−1)3)= (2n+1)3−1.\begin{align*} &\,(3^3-1^3)+(5^3-3^3)\\[4mm] &\,\hspace{2pt}+(7^3-5^3)+\cdots\\[4mm] &\,\hspace{2pt}+\big((2n-3)^3-(2n-5)^3\big)\\[4mm] &\,\hspace{4pt}+\big((2n-1)^3-(2n-3)^3\big)\\[4mm] &\,\hspace{6pt}+\big((2n+1)^3-(2n-1)^3\big)\\[4mm] =&\,(2n+1)^3-1. \end{align*}

The expanded pattern is for sufficiently large nn; the resulting identity holds for all positive nn.

Thus

24∑r=1nr2+2n= (2n+1)3−1,24∑r=1nr2= 8n3+12n2+4n= 4n(n+1)(2n+1).\begin{align*} 24\sum_{r=1}^n r^2+2n=&\,(2n+1)^3-1,\\[4mm] 24\sum_{r=1}^n r^2=&\,8n^3+12n^2+4n\\[4mm] =&\,4n(n+1)(2n+1). \end{align*}

Dividing by 2424 proves

∑r=1nr2=16n(n+1)(2n+1).\boxed{\sum_{r=1}^n r^2=\frac16n(n+1)(2n+1)}.

(b)

解法一

思路

展开

先把每个平方各计一次，再把偶数项额外计两次。偶数平方又可提出4，便能两次调用(a)。

答题过程

展开

The odd squares have weight 11 and the even squares weight 33. Therefore

S2n= ∑r=12nr2+2∑r=1n(2r)2= 2n(2n+1)(4n+1)6 +8n(n+1)(2n+1)6= n(2n+1)3 ×[(4n+1)+4(n+1)]= 13n(2n+1)(8n+5).\begin{align*} S_{2n}=&\,\sum_{r=1}^{2n}r^2+2\sum_{r=1}^n(2r)^2\\[4mm] =&\,\frac{2n(2n+1)(4n+1)}6\\[4mm] &\,+8\frac{n(n+1)(2n+1)}6\\[4mm] =&\,\frac{n(2n+1)}3\\[4mm] &\,\times\big[(4n+1)+4(n+1)\big]\\[4mm] =&\,\boxed{\frac13n(2n+1)(8n+5)}. \end{align*}

Hence a=8,b=5\boxed{a=8,\quad b=5}.

解法二

思路

展开

也可先按奇数项与偶数项成对分组，每一对展开后是r的二次式，再求和。

答题过程

展开

Grouping the terms in pairs,

S2n= ∑r=1n[(2r−1)2+3(2r)2]= ∑r=1n(16r2−4r+1)= 16n(n+1)(2n+1)6 −4n(n+1)2+n= n3[8(n+1)(2n+1) −6(n+1)+3]= n3(16n2+18n+5)= 13n(2n+1)(8n+5).\begin{align*} S_{2n}=&\,\sum_{r=1}^n\big[(2r-1)^2+3(2r)^2\big]\\[4mm] =&\,\sum_{r=1}^n(16r^2-4r+1)\\[4mm] =&\,\frac{16n(n+1)(2n+1)}6\\[4mm] &\,-4\frac{n(n+1)}2+n\\[4mm] =&\,\frac n3\big[8(n+1)(2n+1)\\[4mm] &\,\hspace{2pt}-6(n+1)+3\big]\\[4mm] =&\,\frac n3(16n^2+18n+5)\\[4mm] =&\,\boxed{\frac13n(2n+1)(8n+5)}. \end{align*}

Thus a=8,b=5\boxed{a=8,\quad b=5}.

(c)

解法一

思路

展开

只保留最高次项的系数，或将各因子除以n后取极限。

答题过程

展开

Using (b),

S2nn3= 13(2+1n)(8+5n)⟶ 163.\begin{align*} \frac{S_{2n}}{n^3} =&\,\frac13\bigg(2+\frac1n\bigg)\bigg(8+\frac5n\bigg)\\[4mm] \longrightarrow&\,\boxed{\frac{16}{3}}. \end{align*}