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CIE 9709 2022 November Paper 13 Q3

A Level / CIE / P1

CIE 9709 2022 Nov Paper 13 Paper · Question 3

题目

Problem

(a) Find the first three terms in ascending powers of xx of the expansion of (1+2x)5(1+2x)^5.

(2)

(b) Find the first three terms in ascending powers of xx of the expansion of (1−3x)4(1-3x)^4.

(2)

(c) Hence find the coefficient of x2x^2 in the expansion of (1+2x)5(1−3x)4(1+2x)^5(1-3x)^4.

(2)
题目中文翻译

(a) 求 (1+2x)5(1+2x)^5 按 xx 的升幂展开的前三项。

(2)

(b) 求 (1−3x)4(1-3x)^4 按 xx 的升幂展开的前三项。

(2)

(c) 由此求 (1+2x)5(1−3x)4(1+2x)^5(1-3x)^4 的展开式中 x2x^2 的系数。

(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

选取零个、一个和两个 2x2x,乘以相应的二项式系数。

答题过程

展开

The first three terms are 1+(51)(2x)+(52)(2x)2=1+10x+40x21+\binom51(2x)+\binom52(2x)^2=1+10x+40x^2.

(b)

解法一

思路

展开

同样展开到二次项。一次项为负,二次项为正。

答题过程

展开

The first three terms are 1+(41)(−3x)+(42)(−3x)2=1−12x+54x21+\binom41(-3x)+\binom42(-3x)^2=1-12x+54x^2.

(c)

解法一

思路

展开

只取两个次数相加等于 22 的项来相乘。

答题过程

展开

The coefficient of x2x^2 is 54+10(−12)+40=−2654+10(-12)+40=-26.