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CIE 9709 2022 March Paper 12 Q3

A Level / CIE / P1

CIE 9709 2022 March Paper 12 Paper · Question 3

题目

Problem

Find the term independent of xx in each of the following expansions.

(a) (3x+2x2)6\big(3x+\frac2{x^2}\big)^6

(3)

(b) (3x+2x2)6(1−x3)\big(3x+\frac2{x^2}\big)^6(1-x^3)

(3)
题目中文翻译

求下列各展开式中与 xx 无关的项。

(a) (3x+2x2)6\big(3x+\frac2{x^2}\big)^6

(3)

(b) (3x+2x2)6(1−x3)\big(3x+\frac2{x^2}\big)^6(1-x^3)

(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

每从括号里多选一个 2/x22/x^2,总指数就减少 33。先确定要选几次。

答题过程

展开

Choosing 2/x22/x^2 a total of jj times gives the term (6j)36−j2jx6−3j\binom6j3^{6-j}2^jx^{6-3j}. The constant term uses j=2j=2 and is

(62)3422=4860.\binom62 3^4 2^2=4860.

(b)

解法一

思路

展开

乘以 −x3-x^3 后,原展开式的 x−3x^{-3} 项也会变成常数。用 (a) 的指数规律找它。

答题过程

展开

The original constant contributes 48604860. To obtain an additional constant from the factor −x3-x^3, we need 6−3j=−36-3j=-3, giving j=3j=3. The coefficient of x−3x^{-3} is (63)3323=4320\binom63 3^3 2^3=4320, and it contributes −4320-4320. The required constant is 4860−4320=5404860-4320=540.