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CIE 9231 2026 June Paper 13 Q4

A Level / CIE / FP1

CIE 9231 2026 June Paper 13 Paper · Question 4

题目

Problem

(a) By considering (2r+1)4−(2r−1)4(2r+1)^4-(2r-1)^4, use the method of differences to prove that

∑r=1nr3=14n2(n+1)2.\sum_{r=1}^{n}r^3=\frac{1}{4}n^2(n+1)^2.
(6)

Let Sn=13−23+33−43+…+(−1)n+1n3S_n=1^3-2^3+3^3-4^3+\ldots+(-1)^{n+1}n^3.

(b) Show that S2n=n2(an+b)S_{2n}=n^2(an+b), where aa and bb are integers to be determined.

(3)

题目中文翻译

(a) 通过考虑 (2r+1)4−(2r−1)4(2r+1)^4-(2r-1)^4，使用差分法证明

∑r=1nr3=14n2(n+1)2.\sum_{r=1}^{n}r^3=\frac{1}{4}n^2(n+1)^2.
(6)

设 Sn=13−23+33−43+…+(−1)n+1n3S_n=1^3-2^3+3^3-4^3+\ldots+(-1)^{n+1}n^3。

(b) 证明 S2n=n2(an+b)S_{2n}=n^2(an+b)，其中 aa 和 bb 为待求的整数。

(3)

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

按要求展开指定差式，逐项求和展示相消，再用自然数和消去一次项。

答题过程

展开

Expanding the two powers gives

(2r+1)4= 16r4+32r3+24r2+8r+1,(2r−1)4= 16r4−32r3+24r2−8r+1.\begin{align*} (2r+1)^4=&\,16r^4+32r^3+24r^2+8r+1, \\[4mm] (2r-1)^4=&\,16r^4-32r^3+24r^2-8r+1. \end{align*}

Thus their difference is 64r3+16r64r^3+16r. Write Un=∑r=1nr3U_n=\sum_{r=1}^n r^3 and Vn=∑r=1nrV_n=\sum_{r=1}^n r. For n≥6n\ge6, summing the first and last three terms displays the cancellation below. For 1≤n≤51\le n\le5, directly cancelling the corresponding terms gives the same result.

64Un+16Vn= (34−14)+(54−34) +(74−54)+⋯ +((2n−3)4−(2n−5)4) +((2n−1)4−(2n−3)4) +((2n+1)4−(2n−1)4)= (2n+1)4−1.\begin{align*} 64U_n+16V_n =&\,(3^4-1^4)+(5^4-3^4)\\[4mm] &\,\hspace{2pt}+(7^4-5^4)+\cdots \\[4mm] &\,\hspace{2pt}+\big((2n-3)^4-(2n-5)^4\big) \\[4mm] &\,\hspace{4pt}+\big((2n-1)^4-(2n-3)^4\big) \\[4mm] &\,\hspace{6pt}+\big((2n+1)^4-(2n-1)^4\big) \\[4mm] =&\,(2n+1)^4-1. \end{align*}

The intermediate fourth powers cancel in adjacent terms. Using ∑r=n(n+1)/2\sum r=n(n+1)/2,

64∑r=1nr3= (2n+1)4−1−8n(n+1)= 16n4+32n3+24n2 +8n−8n2−8n= 16n2(n+1)2.\begin{align*} 64\sum_{r=1}^n r^3 =&\,(2n+1)^4-1-8n(n+1) \\[4mm] =&\,16n^4+32n^3+24n^2\\[4mm] &\,+8n-8n^2-8n \\[4mm] =&\,16n^2(n+1)^2. \end{align*}

Dividing by 6464 gives

∑r=1nr3=14n2(n+1)2,\boxed{\sum_{r=1}^n r^3=\frac14n^2(n+1)^2},

as required.

(b)

解法一

思路

展开

先把所有立方和加起来，再扣掉偶数立方和的两倍，使偶数项由正变负。

答题过程

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We can write

S2n=∑r=12nr3−2∑r=1n(2r)3.S_{2n}=\sum_{r=1}^{2n}r^3-2\sum_{r=1}^n(2r)^3.

Using (a),

S2n= 14(2n)2(2n+1)2−1614n2(n+1)2= n2((2n+1)2−4(n+1)2)= n2(4n2+4n+1−4n2−8n−4)= n2(−4n−3).\begin{align*} S_{2n} =&\,\frac14(2n)^2(2n+1)^2 -16\frac14n^2(n+1)^2 \\[4mm] =&\,n^2\big((2n+1)^2-4(n+1)^2\big) \\[4mm] =&\,n^2(4n^2+4n+1-4n^2-8n-4) \\[4mm] =&\,\boxed{n^2(-4n-3)}. \end{align*}

Hence a=−4, b=−3\boxed{a=-4,\ b=-3}.

解法二

思路

展开

也可把每个奇数项与随后的偶数项配对，立方差展开后只需求平方和、自然数和。

答题过程

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Pairing adjacent terms,

S2n= ∑r=1n((2r−1)3−(2r)3)= ∑r=1n(−12r2+6r−1)= −2n(n+1)(2n+1)+3n(n+1)−n= −4n3−6n2−2n+3n2+3n−n= n2(−4n−3).\begin{align*} S_{2n}=&\,\sum_{r=1}^n\big((2r-1)^3-(2r)^3\big) \\[4mm] =&\,\sum_{r=1}^n(-12r^2+6r-1) \\[4mm] =&\,-2n(n+1)(2n+1)+3n(n+1)-n \\[4mm] =&\,-4n^3-6n^2-2n+3n^2+3n-n \\[4mm] =&\,\boxed{n^2(-4n-3)}. \end{align*}

Thus a=−4, b=−3\boxed{a=-4,\ b=-3}.