CalcGospel 國際數學圖譜
返回

CIE 9231 2024 June Paper 13 Q4

A Level / CIE / FP1

CIE 9231 2024 June Paper 13 Paper · Question 4

题目

Problem

(a) Prove by mathematical induction that, for all positive integers nn,

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

The sum SnS_n is defined by Sn=∑r=1nr4S_n=\displaystyle\sum_{r=1}^{n}r^4.

(b) Using the identity

(2r+1)5−(2r−1)5≡160r4+80r2+2,(2r+1)^5-(2r-1)^5\equiv160r^4+80r^2+2,

show that Sn=130n(n+1)(2n+1)(3n2+3n−1)S_n=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1).

(6)

(c) Find the value of lim⁡n→∞(n−5Sn)\displaystyle\lim_{n\to\infty}(n^{-5}S_n).

(2)
题目中文翻译

(a) 用数学归纳法证明：对所有正整数 nn，

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

和 SnS_n 定义为 Sn=∑r=1nr4S_n=\displaystyle\sum_{r=1}^{n}r^4。

(b) 使用恒等式

(2r+1)5−(2r−1)5≡160r4+80r2+2,(2r+1)^5-(2r-1)^5\equiv160r^4+80r^2+2,

证明 Sn=130n(n+1)(2n+1)(3n2+3n−1)S_n=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)。

(6)

(c) 求 lim⁡n→∞(n−5Sn)\displaystyle\lim_{n\to\infty}(n^{-5}S_n) 的值。

(2)
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

归纳时在前k项之和上加第k+1项，再提取k+1并分解二次因子。

答题过程

展开

For n=1n=1, the left-hand side is 11 and the right-hand side is 16(1)(2)(3)=1\frac16(1)(2)(3)=1, so the formula holds.

Assume that, for some positive integer kk,

∑r=1kr2=16k(k+1)(2k+1).\sum_{r=1}^k r^2=\frac16k(k+1)(2k+1).

Then

∑r=1k+1r2= 16k(k+1)(2k+1)+(k+1)2= k+16[k(2k+1)+6(k+1)]= k+16(2k2+7k+6)= 16(k+1)(k+2)(2k+3).\begin{align*} \sum_{r=1}^{k+1}r^2 =&\,\frac16k(k+1)(2k+1)+(k+1)^2\\[4mm] =&\,\frac{k+1}{6}\big[k(2k+1)+6(k+1)\big]\\[4mm] =&\,\frac{k+1}{6}(2k^2+7k+6)\\[4mm] =&\,\frac16(k+1)(k+2)(2k+3). \end{align*}

This is the required formula with n=k+1n=k+1. Since it holds for n=1n=1 and truth for n=kn=k implies truth for n=k+1n=k+1, it follows by induction that the formula holds for every positive integer nn.

(b)

解法一

思路

展开

对恒等式两边求和，左边相邻奇数的五次幂相消，右边包含所求四次和与已证明的平方和。整理时先提取2n+1，减少展开负担。

答题过程

展开

Summing the given identity from r=1r=1 to r=nr=n, the left-hand side telescopes:

 (35−15)+(55−35) +(75−55)+⋯ +((2n−3)5−(2n−5)5) +((2n−1)5−(2n−3)5) +((2n+1)5−(2n−1)5)= (2n+1)5−1.\begin{align*} &\,(3^5-1^5)+(5^5-3^5)\\[4mm] &\,\hspace{2pt}+(7^5-5^5)+\cdots\\[4mm] &\,\hspace{2pt}+\big((2n-3)^5-(2n-5)^5\big)\\[4mm] &\,\hspace{4pt}+\big((2n-1)^5-(2n-3)^5\big)\\[4mm] &\,\hspace{6pt}+\big((2n+1)^5-(2n-1)^5\big)\\[4mm] =&\,(2n+1)^5-1. \end{align*}

This display illustrates the cancellation for sufficiently large nn; the resulting identity holds for all positive nn.

The sum of the right-hand side, using (a), is

 160Sn+80∑r=1nr2+2n= 160Sn+403n(n+1)(2n+1)+2n.\begin{align*} &\,160S_n+80\sum_{r=1}^n r^2+2n\\[4mm] =&\,160S_n+\frac{40}{3}n(n+1)(2n+1)+2n. \end{align*}

Therefore

160Sn= (2n+1)5−1 −403n(n+1)(2n+1)−2n= (2n+1)[(2n+1)4−1−403n(n+1)]= (2n+1)[16n4+32n3 +323n2−163n]= 163n(2n+1)(3n3+6n2+2n−1)= 163n(n+1)(2n+1)(3n2+3n−1).\begin{align*} 160S_n =&\,(2n+1)^5-1\\[4mm] &\,-\frac{40}{3}n(n+1)(2n+1)-2n\\[4mm] =&\,(2n+1)\big[(2n+1)^4-1-\tfrac{40}{3}n(n+1)\big]\\[4mm] =&\,(2n+1)\big[16n^4+32n^3\\[4mm] &\,\hspace{2pt}+\tfrac{32}{3}n^2-\tfrac{16}{3}n\big]\\[4mm] =&\,\frac{16}{3}n(2n+1)(3n^3+6n^2+2n-1)\\[4mm] =&\,\frac{16}{3}n(n+1)(2n+1)(3n^2+3n-1). \end{align*}

Dividing by 160160 gives

Sn=130n(n+1)(2n+1)(3n2+3n−1),\boxed{S_n=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)},

as required.

(c)

解法一

思路

展开

把因式分解式除以n五次，每一因子提出相应的n，剩余括号逐项取极限。

答题过程

展开

By (b),

n−5Sn=130(1+1n)(2+1n)(3+3n−1n2).n^{-5}S_n=\frac1{30}\bigg(1+\frac1n\bigg) \bigg(2+\frac1n\bigg) \bigg(3+\frac3n-\frac1{n^2}\bigg).

Hence

lim⁡n→∞(n−5Sn)=130(1)(2)(3)=15.\boxed{\lim_{n\to\infty}(n^{-5}S_n) =\frac1{30}(1)(2)(3)=\frac15}.