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CIE 9709 2024 November Paper 12 Q4

A Level / CIE / P1

CIE 9709 2024 Nov Paper 12 Paper · Question 4

题目

Problem

Find the term independent of xx in the expansion of each of the following:

(a) (x+3x2)6\left(x+\dfrac{3}{x^2}\right)^6

[2]

(b) (4x3−5)(x+3x2)6(4x^3-5)\left(x+\dfrac{3}{x^2}\right)^6.

[4]
题目中文翻译

求下列各式展开式中与 xx 无关的项:

(a) (x+3x2)6\left(x+\dfrac{3}{x^2}\right)^6

(b) (4x3−5)(x+3x2)6(4x^3-5)\left(x+\dfrac{3}{x^2}\right)^6。

解答

(a)

解法一

思路

展开

若从六个因子中选 kk 个 3/x23/x^2,总幂次为 6−3k6-3k。令它等于零便得到常数项。

答题过程

展开

The general term is (6k)3kx6−3k\binom6k 3^k x^{6-3k}. It is constant when k=2k=2, giving (62)32=135\binom62 3^2=135.

(b)

解法一

思路

展开

外面的 −5-5 要配常数项,4x34x^3 则要配原展开式中的 x−3x^{-3} 项。

答题过程

展开

The term in x−3x^{-3} occurs when 6−3k=−36-3k=-3, so k=3k=3. Its coefficient is (63)33=540\binom63 3^3=540. Using the constant term from (a), the required term is

4(540)−5(135)=1485.4(540)-5(135)=1485.